1$ yields fewer than 199 divisible numbers (specifically, at most 198), it is excluded by the problem hypothesis.\\nThus, the only valid state is where all 200 integers are divisible.\\n\\nQ.E.D.\\n\\nThe final result is that all 200 numbers must be divisible by 2013.\\n\\nFinal Answer: All 200 of the integers are divisible by 2013.\\n\\n## Attempt 2\\nThe problem asks us to prove that if at least 199 out of 200 integers (formed by rows and columns of a $100 \\\\times 100$ table of nonzero digits) are divisible by $2013$, then all 200 must be divisible by $2013$.\\n\\nLet $A = (a_{i,j})$ be the table with $1 \\\\le i, j \\\\le 100$, where $a_{i,j} \\\\in \\\\{1, 2, \\\\dots, 9\\\\}$.\\nLet $R_i$ be the integer formed by the $i$-th row: $R_i = \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j}$.\\nLet $C_j$ be the integer formed by the $j$-th column: $C_j = \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\\nWe have a set of 200 integers $\\\\mathcal{N} = \\\\{R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}\\\\}$.\\nThe condition is that at least 199 of these integers are divisible by $m = 2013$.\\nThis is equivalent to saying that there is at most one integer in $\\\\mathcal{N}$ that is not divisible by $m$.\\nLet $S$ be the set of indices corresponding to the integers not divisible by $m$. The condition is $|S| \\\\le 1$.\\nIf $|S| = 0$, then all 200 integers are divisible by $m$, and the statement holds.\\nWe proceed to show that the case $|S| = 1$ leads to a contradiction.\\nLet $X$ be the unique integer in $\\\\mathcal{N}$ that is not divisible by $m$.\\nLet us denote $r_i = R_i \\\\pmod m$ and $c_j = C_j \\\\pmod m$. The assumption is that exactly one of these values is non-zero.\\n\\nFirst, factorize $m$:\\n$$ 2013 = 3 \\\\times 671 = 3 \\\\times 11 \\\\times 61 $$\\nLet $p$ be a prime divisor of $m$. Since $m$ divides an integer implies divisibility by $p$, if a number is divisible by $m$, it is divisible by $p$. Conversely, if a number is not divisible by $m$, it might or might not be divisible by $p$. However, if $X$ is not divisible by $m$, it is not divisible by at least one prime factor of $m$.\\nSpecifically, if we can prove that $X$ is divisible by 3, 11, and 61, then $X$ must be divisible by $3 \\\\times 11 \\\\times 61 = 2013$, which contradicts the assumption that $X$ is not divisible by $2013$. Thus, we aim to show that $X$ must be divisible by each prime factor $p \\\\in \\\\{3, 11, 61\\\\}$.\\n\\n**Part 1: Divisibility by 3**\\nAssume $X$ is the unique number not divisible by $m$. Then $X$ is the unique number in $\\\\mathcal{N}$ not divisible by $m$.\\nIf $X$ is not divisible by 3, then $X$ is the unique number not divisible by 3.\\nLet $u_i = \\\\sum_{j=1}^{100} a_{i,j} \\\\pmod 3$ and $v_j = \\\\sum_{i=1}^{100} a_{i,j} \\\\pmod 3$.\\nNote that $R_i \\\\equiv \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j} \\\\equiv \\\\sum_{j=1}^{100} a_{i,j} \\\\pmod 3$ since $10 \\\\equiv 1 \\\\pmod 3$.\\nSimilarly, $C_j \\\\equiv \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i} \\\\equiv \\\\sum_{i=1}^{100} a_{i,j} \\\\pmod 3$.\\nLet $S_R = \\\\sum_{i=1}^{100} R_i \\\\pmod 3$ and $S_C = \\\\sum_{j=1}^{100} C_j \\\\pmod 3$.\\nCalculating the sums:\\n$$ S_R = \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j} \\\\equiv \\\\sum_{i,j} a_{i,j} \\\\pmod 3 $$\\n$$ S_C = \\\\sum_{j=1}^{100} \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i} \\\\equiv \\\\sum_{i,j} a_{i,j} \\\\pmod 3 $$\\nThus, $S_R \\\\equiv S_C \\\\pmod 3$.\\nIf $X = R_k$ is the unique exception, then $R_k \\\\not\\\\equiv 0 \\\\pmod 3$ and all other $R_i, C_j$ are divisible by $m$, hence by 3. Thus all $C_j \\\\equiv 0 \\\\pmod 3$.\\nThen $S_R \\\\equiv R_k \\\\not\\\\equiv 0 \\\\pmod 3$.\\nHowever, $S_C \\\\equiv 0 \\\\pmod 3$ (since all $C_j \\\\equiv 0$).\\nThis contradicts $S_R \\\\equiv S_C \\\\pmod 3$.\\nSimilarly, if $X = C_k$ is the unique exception, we get a contradiction.\\nTherefore, the unique exception $X$ must be divisible by 3.\\n\\n**Part 2: Divisibility by 11**\\nAssume $X$ is the unique number not divisible by $m$. We work modulo 11, where $10 \\\\equiv -1 \\\\pmod{11}$.\\nWe establish an identity relating the row values and column values.\\nConsider $\\\\Sigma_R = \\\\sum_{i=1}^{100} R_i (-1)^i \\\\pmod{11}$ and $\\\\Sigma_C = \\\\sum_{j=1}^{100} C_j (-1)^j \\\\pmod{11}$.\\n$$ \\\\Sigma_R = \\\\sum_{i=1}^{100} \\\\left( \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j} \\\\right) (-1)^i \\\\equiv \\\\sum_{i,j} a_{i,j} (-1)^{100-j} (-1)^i \\\\pmod{11} $$\\nSince $(-1)^{100} = 1$, $(-1)^{100-j} = (-1)^{-j} = (-1)^j$.\\nSo $\\\\Sigma_R \\\\equiv \\\\sum_{i,j} a_{i,j} (-1)^{i+j} \\\\pmod{11}$.\\nSimilarly,\\n$$ \\\\Sigma_C = \\\\sum_{j=1}^{100} \\\\left( \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i} \\\\right) (-1)^j \\\\equiv \\\\sum_{i,j} a_{i,j} (-1)^{100-i} (-1)^j \\\\equiv \\\\sum_{i,j} a_{i,j} (-1)^{i+j} \\\\pmod{11} $$\\nThus, $\\\\Sigma_R \\\\equiv \\\\Sigma_C \\\\pmod{11}$.\\nSuppose $X = R_k$ is the unique exception. Then all $C_j$ are divisible by $m$, so $C_j \\\\equiv 0 \\\\pmod{11}$ for all $j$.\\nThis implies $\\\\Sigma_C = \\\\sum C_j (-1)^j \\\\equiv 0 \\\\pmod{11}$.\\nConsequently, $\\\\Sigma_R \\\\equiv 0 \\\\pmod{11}$.\\nAlso, for all $i \\\\ne k$, $R_i$ is divisible by $m$, so $R_i \\\\equiv 0 \\\\pmod{11}$.\\nThus, $\\\\Sigma_R = \\\\sum_{i=1}^{100} R_i (-1)^i \\\\equiv R_k (-1)^k \\\\pmod{11}$.\\nFrom $\\\\Sigma_R \\\\equiv 0$, we get $R_k (-1)^k \\\\equiv 0 \\\\pmod{11}$. Since $(-1)^k \\\\not\\\\equiv 0$, we have $R_k \\\\equiv 0 \\\\pmod{11}$.\\nIf $X = C_k$ is the unique exception, the same logic yields $C_k \\\\equiv 0 \\\\pmod{11}$.\\nThus, the unique exception $X$ must be divisible by 11.\\n\\n**Part 3: Divisibility by 61**\\nAssume $X$ is the unique number not divisible by $m$. We work modulo 61. Note that $\\\\gcd(10, 61) = 1$, so 10 is invertible modulo 61. Let $10^{-1}$ denote the multiplicative inverse of 10 modulo 61.\\nConsider $\\\\Phi_R = \\\\sum_{i=1}^{100} R_i (10^{-1})^i$ and $\\\\Phi_C = \\\\sum_{j=1}^{100} C_j (10^{-1})^j$.\\n$$ \\\\Phi_R = \\\\sum_{i=1}^{100} \\\\left( \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j} \\\\right) 10^{-i} = \\\\sum_{i,j} a_{i,j} 10^{100-i-j} $$\\n$$ \\\\Phi_C = \\\\sum_{j=1}^{100} \\\\left( \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i} \\\\right) 10^{-j} = \\\\sum_{i,j} a_{i,j} 10^{100-i-j} $$\\nThus, $\\\\Phi_R = \\\\Phi_C$ as rational numbers, and consequently $\\\\Phi_R \\\\equiv \\\\Phi_C \\\\pmod{61}$.\\nSuppose $X = R_k$ is the unique exception. Then all $C_j$ are divisible by $m$, so $C_j \\\\equiv 0 \\\\pmod{61}$ for all $j$.\\nThus $\\\\Phi_C \\\\equiv 0 \\\\pmod{61}$.\\nThis implies $\\\\Phi_R \\\\equiv 0 \\\\pmod{61}$.\\nFor all $i \\\\ne k$, $R_i$ is divisible by $m$, so $R_i \\\\equiv 0 \\\\pmod{61}$.\\nThus $\\\\Phi_R = \\\\sum_{i=1}^{100} R_i 10^{-i} \\\\equiv R_k 10^{-k} \\\\pmod{61}$.\\nFrom $\\\\Phi_R \\\\equiv 0$, we get $R_k 10^{-k} \\\\equiv 0 \\\\pmod{61}$. Since 10 is invertible, $R_k \\\\equiv 0 \\\\pmod{61}$.\\nSimilarly, if $X = C_k$ is the unique exception, $C_k \\\\equiv 0 \\\\pmod{61}$.\\nThus, the unique exception $X$ must be divisible by 61.\\n\\n**Conclusion**\\nWe have shown that if there is a unique number $X$ among the 200 integers that is not divisible by 2013, then $X$ must be divisible by 3, 11, and 61. Therefore, $X$ must be divisible by $3 \\\\times 11 \\\\times 61 = 2013$.\\nThis contradicts the assumption that $X$ is not divisible by 2013.\\nHence, it is impossible for exactly 199 numbers to be divisible by 2013.\\nTherefore, if at least 199 numbers are divisible by 2013, all 200 numbers must be divisible by 2013.\\n\\nFinal Answer: All of the 200 integers are divisible by 2013.\\n\\n## Attempt 3\\nThe problem asks us to prove that if at least 199 of the 200 integers (formed by the rows and columns of a $100 \\\\times 100$ grid of nonzero digits) are divisible by $2013$, then all 200 integers are divisible by $2013$.\\n\\nLet the entries of the grid be $x_{i,j} \\\\in \\\\{1, 2, \\\\dots, 9\\\\}$ for $1 \\\\le i, j \\\\le 100$.\\nThe row integer $R_i$ is formed by the digits $x_{i,1}, x_{i,2}, \\\\dots, x_{i,100}$ read from left to right. Thus,\\n$$ R_i = \\\\sum_{j=1}^{100} x_{i,j} 10^{100-j}. $$\\nThe column integer $C_j$ is formed by the digits $x_{1,j}, x_{2,j}, \\\\dots, x_{100,j}$ read from top to bottom. Thus,\\n$$ C_j = \\\\sum_{i=1}^{100} x_{i,j} 10^{100-i}. $$\\nWe are given that at least 199 of these 200 numbers ($R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}$) are divisible by $2013$.\\nWe want to show that all 200 are divisible by $2013$.\\nIt suffices to show that it is impossible for exactly one of the numbers to be not divisible by $2013$.\\n\\nFirst, observe the properties of the modulus $2013$.\\n$2013 = 3 \\\\times 671 = 3 \\\\times 11 \\\\times 61$.\\nSince $2013$ is coprime to $10$ ($2013$ is not divisible by 2 or 5), $10$ is invertible modulo $2013$. This implies that $10^k$ is invertible modulo $2013$ for any integer $k \\\\ge 0$.\\n\\nSuppose, for the sake of contradiction, that exactly one number is not divisible by $2013$. There are two cases for the location of this exceptional number: it is either a row integer $R_k$ or a column integer $C_l$.\\n\\n**Case 1:** $R_k$ is the unique number not divisible by $2013$.\\nThis implies that all columns $C_1, \\\\dots, C_{100}$ are divisible by $2013$, and all other row integers $R_i$ (for $i \\\\ne k$) are divisible by $2013$.\\nThe condition that all columns are divisible by $2013$ means:\\n$$ \\\\sum_{i=1}^{100} x_{i,j} 10^{100-i} \\\\equiv 0 \\\\pmod{2013} \\\\quad \\\\text{for each } j=1, \\\\dots, 100. $$\\nLet $N=100$. The congruence can be written as:\\n$$ x_{N,j} + \\\\sum_{i=1}^{N-1} x_{i,j} 10^{N-i} \\\\equiv 0 \\\\pmod{2013}. $$\\nSince the coefficient of $x_{N,j}$ is $10^0 = 1$, we can express $x_{N,j}$ as:\\n$$ x_{N,j} \\\\equiv - \\\\sum_{i=1}^{N-1} x_{i,j} 10^{N-i} \\\\pmod{2013}. $$\\nNow consider the row integer $R_N$. By definition:\\n$$ R_N = \\\\sum_{j=1}^{N} x_{N,j} 10^{N-j}. $$\\nSubstituting the expression for $x_{N,j}$:\\n$$ R_N \\\\equiv \\\\sum_{j=1}^{N} \\\\left( - \\\\sum_{i=1}^{N-1} x_{i,j} 10^{N-i} \\\\right) 10^{N-j} \\\\pmod{2013}. $$\\nRearranging the summation order:\\n$$ R_N \\\\equiv - \\\\sum_{i=1}^{N-1} 10^{N-i} \\\\left( \\\\sum_{j=1}^{N} x_{i,j} 10^{N-j} \\\\right) \\\\pmod{2013}. $$\\nNotice that the inner sum $\\\\sum_{j=1}^{N} x_{i,j} 10^{N-j}$ is precisely the integer $R_i$. Thus:\\n$$ R_N \\\\equiv - \\\\sum_{i=1}^{N-1} 10^{N-i} R_i \\\\pmod{2013}. $$\\nWe analyze the consequences of this identity under our assumption that exactly one row is non-divisible.\\n\\n* **Subcase 1.1:** $k = N$.\\n The assumption is that $R_1, \\\\dots, R_{N-1}$ are all divisible by $2013$ (i.e., $R_i \\\\equiv 0$ for $i < N$).\\n Substituting into the identity:\\n $$ R_N \\\\equiv - \\\\sum_{i=1}^{N-1} 10^{N-i} \\\\cdot 0 \\\\equiv 0 \\\\pmod{2013}. $$\\n This contradicts the assumption that $R_N$ is not divisible by $2013$.\\n\\n* **Subcase 1.2:** $k < N$.\\n The assumption is that $R_N$ is divisible by $2013$ (since it's not the exceptional row).\\n The identity becomes:\\n $$ 0 \\\\equiv - \\\\sum_{i=1}^{N-1} 10^{N-i} R_i \\\\pmod{2013}. $$\\n Since only $R_k$ is not zero (and equals $R_k \\\\pmod{2013}$), all other $R_i$ (for $i < N, i \\\\ne k$) are $0$. The sum reduces to:\\n $$ 0 \\\\equiv - 10^{N-k} R_k \\\\pmod{2013}. $$\\n Since $2013$ is coprime to $10$, $10^{N-k}$ is invertible modulo $2013$. Multiplying by the inverse of $10^{N-k}$ yields:\\n $$ R_k \\\\equiv 0 \\\\pmod{2013}. $$\\n This contradicts the assumption that $R_k$ is not divisible by $2013$.\\n\\nSince both subcases lead to a contradiction, the assumption that $R_k$ is the unique non-divisible number is false. Thus, if all columns are divisible, all rows must be divisible.\\n\\n**Case 2:** $C_l$ is the unique number not divisible by $2013$.\\nThis implies that all row integers $R_1, \\\\dots, R_{100}$ are divisible by $2013$, and all other column integers $C_j$ (for $j \\\\ne l$) are divisible.\\nConsider the transpose matrix $A^T$. The rows of $A^T$ correspond to the columns of $A$, and the columns of $A^T$ correspond to the rows of $A$.\\nThe condition \\\"all rows $R_i$ are divisible\\\" translates to \\\"all columns of $A^T$ are divisible\\\".\\nThe condition \\\"$C_j$ are divisible for $j \\\\ne l$\\\" translates to \\\"all rows of $A^T$ are divisible, except possibly the $l$-th row\\\".\\nFrom the symmetry of the argument (or simply repeating Case 1 with roles swapped), if all columns of a matrix are divisible, then all rows must be divisible. Applying this to $A^T$, we conclude that all rows of $A^T$ are divisible.\\nSince the rows of $A^T$ are the columns of $A$, this implies that all columns $C_j$ are divisible by $2013$.\\nThis contradicts the assumption that $C_l$ is not divisible by $2013$.\\n\\n**Conclusion:**\\nIn both possible cases for a single non-divisible number, we arrive at a contradiction. Therefore, it is impossible for exactly one of the 200 integers to be non-divisible. Since we are given that at least 199 are divisible, the set of non-divisible integers cannot have cardinality 1. The only remaining option is that the set of non-divisible integers is empty.\\nThus, all 200 numbers must be divisible by $2013$.\\n\\nQ.E.D.\\n\\n## Attempt 4\\nLet $a_{i,j}$ be the nonzero digit in the cell at row $i$ and column $j$, where $1 \\\\le i, j \\\\le 100$.\\nThe $i$-th row forms a $100$-digit integer $R_i$, which can be expressed as:\\n$$R_i = \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j}$$\\nThe $j$-th column forms a $100$-digit integer $C_j$, which can be expressed as:\\n$$C_j = \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i}$$\\nWe consider these values modulo $2013$. Note that $2013 = 3 \\\\times 11 \\\\times 61$.\\nWe observe that $10$ is relatively prime to $2013$ since $\\\\gcd(10, 3) = 1$, $\\\\gcd(10, 11) = 1$, and $\\\\gcd(10, 61) = 1$. Consequently, any power of $10$ is also invertible modulo $2013$. Specifically, for any $k \\\\in \\\\{1, \\\\dots, 100\\\\}$, $w_k = 10^{100-k}$ is coprime to $2013$.\\n\\nConsider the quantity $S$ defined by the double sum of the products of the digits and the appropriate powers of 10:\\n$$S = \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{i,j} 10^{100-i} 10^{100-j}$$\\nWe can evaluate this sum in two different ways.\\nFirst, grouping the terms by columns $j$:\\n$$S = \\\\sum_{j=1}^{100} \\\\left( \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i} \\\\right) 10^{100-j} = \\\\sum_{j=1}^{100} C_j 10^{100-j}$$\\nSecond, grouping the terms by rows $i$:\\n$$S = \\\\sum_{i=1}^{100} \\\\left( \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j} \\\\right) 10^{100-i} = \\\\sum_{i=1}^{100} R_i 10^{100-i}$$\\n\\nWe are given that at least $199$ of the $200$ integers $\\\\{R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}\\\\}$ are divisible by $2013$.\\nThis implies that the set of non-divisible integers has a size of either $0$ or $1$.\\nIf the set is empty, all numbers are divisible, and the claim is proven.\\nSuppose, for the sake of contradiction, that the set of non-divisible integers has size $1$.\\nThere are two cases depending on which number is not divisible.\\n\\n**Case 1:** The number $R_k$ is the unique integer not divisible by $2013$.\\nThis implies that all other rows $R_i$ for $i \\\\neq k$ and all columns $C_j$ for all $j$ are divisible by $2013$.\\nSubstituting the divisibility conditions into our expressions for $S$ modulo $2013$:\\n1. Using the column sum: Since $C_j \\\\equiv 0 \\\\pmod{2013}$ for all $j=1, \\\\dots, 100$, we have\\n $$S = \\\\sum_{j=1}^{100} C_j 10^{100-j} \\\\equiv \\\\sum_{j=1}^{100} 0 \\\\cdot 10^{100-j} \\\\equiv 0 \\\\pmod{2013}$$\\n2. Using the row sum: Since $R_i \\\\equiv 0 \\\\pmod{2013}$ for all $i \\\\neq k$, we have\\n $$S = \\\\sum_{i=1}^{100} R_i 10^{100-i} = R_k 10^{100-k} + \\\\sum_{i \\\\neq k} R_i 10^{100-i} \\\\equiv R_k 10^{100-k} \\\\pmod{2013}$$\\nComparing the two results, we get $R_k 10^{100-k} \\\\equiv 0 \\\\pmod{2013}$.\\nSince $10^{100-k}$ is coprime to $2013$ (as established earlier), it is invertible modulo $2013$. Multiplying by the modular inverse of $10^{100-k}$, we obtain $R_k \\\\equiv 0 \\\\pmod{2013}$.\\nThis contradicts the assumption that $R_k$ is not divisible by $2013$.\\n\\n**Case 2:** The number $C_j$ is the unique integer not divisible by $2013$.\\nBy symmetry, the roles of rows and columns are interchangeable (considering the transpose of the matrix). If $C_j$ is the only non-divisible number, then all $C_i$ ($i \\\\neq j$) and all $R_i$ are divisible by $2013$.\\nWe can apply the same logic. Grouping $S$ by columns, the condition $C_m \\\\equiv 0$ for $m \\\\neq j$ leads to $S \\\\equiv C_j 10^{100-j} \\\\pmod{2013}$.\\nGrouping $S$ by rows, all $R_i$ are divisible by $2013$, so $S \\\\equiv 0 \\\\pmod{2013}$.\\nThis implies $C_j 10^{100-j} \\\\equiv 0 \\\\pmod{2013}$, which implies $C_j \\\\equiv 0 \\\\pmod{2013}$ due to invertibility. This contradicts the assumption that $C_j$ is not divisible.\\n\\nSince both cases lead to a contradiction, it is impossible to have exactly one integer that is not divisible by $2013$.\\nTherefore, if at least $199$ integers are divisible by $2013$, it must be that all $200$ integers are divisible by $2013$.\\n\\n**Final Answer:**\\nAll $200$ numbers (the row and column integers) are divisible by $2013$.\\n\\n# Fixed response prefix\\nLet the table be denoted by $A = (a_{ij})$ for $1 \\\\le i, j \\\\le 100$, where $a_{ij} \\\\in \\\\{0, 1, \\\\dots, 9\\\\}$? The problem says \\\"nonzero digits\\\", so $a_{ij} \\\\in \\\\{1, \\\\dots, 9\\\\}$.\\nWait, usually \\\"digits\\\" implies $\\\\{0, \\\\dots, 9\\\\}$. \\\"Nonzero digits\\\" means $a_{ij} \\\\in \\\\{1, \\\\dots, 9\\\\}$.\\nHowever, this constraint might not be crucial for the modularity of the rows and columns. Let's check if it matters. Usually, divisibility problems depend on sums of digits or alternating sums etc. Here we are dealing with integers formed by concatenating digits.\\nLet $N(i)$ be the integer formed by the $i$-th row. $N(i) = \\\\sum_{j=1}^{100} a_{ij} 10^{100-j}$.\\nLet $C(j)$ be the integer formed by the $j$-th column. $C(j) = \\\\sum_{i=1}^{100} a_{ij} 10^{100-i}$.\\nWe are given that at least 199 of these 200 numbers are divisible by $m = 2013$. We need to prove all of them are divisible by $m$.\\n\\nFirst, let's factorize 2013.\\n$2013 = 3 \\\\times 671$.\\nIs 671 prime?\\n$\\\\sqrt{671} \\\\approx 25.9$.\\nPrimes to check: 2, 3, 5, 7, 11, 13, 17, 19, 23.\\nNot div by 2 (odd).\\nSum of digits $6+7+1=14$, not div by 3.\\nEnds in 1, not div by 5.\\n$671 = 7 \\\\times 95 + 6$, no.\\n$671 = 11 \\\\times 61$. Yes!\\nSo $2013 = 3 \\\\times 11 \\\\times 61$.\\nSince $3, 11, 61$ are pairwise coprime, the condition \\\"$x$ is divisible by 2013\\\" is equivalent to \\\"$x$ is divisible by 3, 11, and 61\\\".\\n\\nLet's denote the condition $P(A)$ as \\\"all rows and columns are divisible by 2013\\\".\\nWe are given that at least 199 out of 200 conditions hold.\\nThis means there are at most 1 exception among the set $\\\\{N(1), \\\\dots, N(100)\\\\} \\\\cup \\\\{C(1), \\\\dots, C(100)\\\\}$.\\nSuppose exactly one number is NOT divisible by 2013.\\nLet's denote $r_i = N(i) \\\\pmod m$ and $c_j = C(j) \\\\pmod m$.\\nThe values $a_{ij}$ are coefficients. Specifically, $N(i) = \\\\sum_{j=1}^{100} a_{ij} B_j^{row}$ and $C(j) = \\\\sum_{i=1}^{100} a_{ij} D_i^{col}$, where $B$ and $D$ are base powers.\\nSpecifically, $N(i) \\\\equiv \\\\sum_{j=1}^{100} a_{ij} (-1)^j \\\\pmod 9$ isn't right because the base is 10.\\nActually, $N(i) = \\\\sum_{j=1}^{100} a_{ij} 10^{100-j}$. Let's re-index powers.\\n$N(i) = \\\\sum_{k=0}^{99} a_{i, 100-k} 10^k$.\\nLet $x_k$ be the value of the column power $10^k$.\\nWait, the problem structure suggests looking at the relation between rows and columns modulo small primes like 3, 11, 61. Or maybe directly 2013?\\nUsually, divisibility by 3 depends on sum of digits.\\n$N(i) \\\\equiv \\\\sum_{j=1}^{100} a_{ij} \\\\pmod 3$.\\n$C(j) \\\\equiv \\\\sum_{i=1}^{100} a_{ij} \\\\pmod 3$.\\nIf all rows are divisible by 3, then $\\\\sum_{j} a_{ij} \\\\equiv 0 \\\\pmod 3$ for all $i$. Summing over $i$, total sum $S = \\\\sum_{i,j} a_{ij} \\\\equiv 0 \\\\pmod 3$.\\nAlso if all columns are divisible by 3, then $\\\\sum_{i} a_{ij} \\\\equiv 0 \\\\pmod 3$ for all $j$.\\nIf all but one are divisible by 3.\\nSuppose $N(i)$ are all divisible by 3 except possibly one. And $C(j)$ are all divisible by 3 except possibly one.\\nIf $N(1), \\\\dots, N(100), C(1), \\\\dots, C(100)$ are all divisible by 3.\\nThen $\\\\sum_{j=1}^{100} a_{ij} \\\\equiv 0 \\\\pmod 3$ for all $i$.\\nAnd $\\\\sum_{i=1}^{100} a_{ij} \\\\equiv 0 \\\\pmod 3$ for all $j$.\\nTotal sum is 0. This is consistent.\\n\\nLet's consider the modulus $m=2013$.\\nIt's likely useful to work with modular arithmetic properties.\\nLet $r_i$ be remainder of row $i$ mod $m$.\\nLet $c_j$ be remainder of col $j$ mod $m$.\\nWe have constraints involving these $r_i$ and $c_j$.\\nNote that the matrix entries $a_{ij}$ are fixed.\\nConsider the linear combination of rows.\\nLet $L = \\\\text{lcm}(1, \\\\dots, 100)$? No.\\nLet's look at the relation between row remainders and column remainders.\\nThis looks like a system of linear equations over $\\\\mathbb{Z}_m$ or something similar. But the variables $a_{ij}$ are not unknowns; they are digits.\\nBut we don't know $a_{ij}$. We just know counts.\\nWait, the structure is rigid.\\nRow $i$: digits $a_{i1}, \\\\dots, a_{i100}$. Value $R_i = \\\\sum a_{ij} 10^{100-j}$.\\nCol $j$: digits $a_{1j}, \\\\dots, a_{100,j}$. Value $K_j = \\\\sum a_{ij} 10^{100-i}$.\\nNotice that if we sum all row values weighted by some coefficients $\\\\alpha_i$, we get $\\\\sum_i \\\\alpha_i R_i = \\\\sum_{i,j} \\\\alpha_i a_{ij} 10^{100-j}$.\\nIf we sum all column values weighted by $\\\\beta_j$, we get $\\\\sum_j \\\\beta_j K_j = \\\\sum_{i,j} \\\\beta_j a_{ij} 10^{100-i}$.\\nFor this to be useful, we want to extract relations between $R_i$ and $K_j$.\\nMaybe choosing weights such that the powers of 10 match?\\nBut powers of 10 for rows are decreasing $10^{99}, 10^{98}, \\\\dots$. For columns they are decreasing $10^{99}, 10^{98}, \\\\dots$.\\nLet's rewrite $R_i = \\\\sum_{j=1}^{100} a_{ij} 10^{100-j}$. Let $P_j = 10^{100-j}$. Then $R_i = \\\\sum_{j=1}^{100} a_{ij} P_j$.\\nHere indices $j$ run $1..100$. $P_1 = 10^{99}, P_{100} = 10^0 = 1$.\\nSimilarly $K_j = \\\\sum_{i=1}^{100} a_{ij} 10^{100-i}$.\\nLet $Q_i = 10^{100-i}$. Then $K_j = \\\\sum_{i=1}^{100} a_{ij} Q_i$.\\nNotice $Q_i$ has the same sequence of values as $P_j$.\\n$Q_1 = P_1, \\\\dots, Q_{100} = P_{100}$.\\nLet's denote $\\\\omega_j = 10^{100-j}$ for $j=1,\\\\dots,100$.\\nThen $R_i = \\\\sum_{j=1}^{100} a_{ij} \\\\omega_j$ and $K_j = \\\\sum_{i=1}^{100} a_{ij} \\\\omega_i$.\\nWait, the powers are different.\\nIn $K_j$, the term $a_{ij}$ corresponds to digit at position $100-i$ from right. In $R_i$, the term $a_{ij}$ corresponds to digit at position $100-j$ from right.\\nAh, in a row, $a_{i1}$ is the leading digit (position 99, i.e., $10^{99}$). $a_{i100}$ is unit digit ($10^0$).\\nIn a column, $a_{1j}$ is the leading digit ($10^{99}$). $a_{100,j}$ is unit digit ($10^0$).\\nSo yes, the positions correspond exactly. The entry $a_{ij}$ is the coefficient of $10^{99}$ in row $i$? No.\\nRow $i$: $a_{i1} a_{i2} \\\\dots a_{i100}$. Value $\\\\sum_{j=1}^{100} a_{ij} 10^{100-j}$.\\nColumn $j$: $a_{1j} a_{2j} \\\\dots a_{100j}$. Value $\\\\sum_{i=1}^{100} a_{ij} 10^{100-i}$.\\nLet $\\\\lambda_k = 10^{100-k}$ for $k=1,\\\\dots,100$.\\nThen $R_i = \\\\sum_{j=1}^{100} a_{ij} \\\\lambda_j$.\\nAnd $K_j = \\\\sum_{i=1}^{100} a_{ij} \\\\lambda_i$.\\nThis is interesting. The set of multipliers used in row calculation is $\\\\{\\\\lambda_1, \\\\dots, \\\\lambda_{100}\\\\}$.\\nThe set of multipliers used in column calculation is $\\\\{\\\\lambda_1, \\\\dots, \\\\lambda_{100}\\\\}$.\\nHowever, in $K_j$, $\\\\lambda_i$ goes with $a_{ij}$.\\nSo $K_j$ uses multipliers $\\\\lambda_i$ for row index $i$.\\n$R_i$ uses multipliers $\\\\lambda_j$ for column index $j$.\\nLet's define $M = 2013$.\\nWe want to show that if $R_i \\\\equiv 0 \\\\pmod M$ for $i \\\\notin I_R$ and $K_j \\\\equiv 0 \\\\pmod M$ for $j \\\\notin I_C$, with $|I_R| + |I_C| \\\\le 1$, then $I_R = \\\\emptyset$ and $I_C = \\\\emptyset$.\\nOr rather, assume one fails. Say $N(100)$ is not divisible, others are.\\nOr any single failure.\\n\\nConsider the sum $S = \\\\sum_{j=1}^{100} (\\\\text{something}) R_j$.\\nLet's try to relate $R$'s and $K$'s using the identity $\\\\sum a_{ij} x^j y^i$ type things? No.\\nConsider the relation $\\\\sum_{i=1}^{100} \\\\mu_i K_i = \\\\sum_{i=1}^{100} \\\\mu_i \\\\sum_{j=1}^{100} a_{ij} \\\\lambda_j$. Wait.\\n$K_i = \\\\sum_{k=1}^{100} a_{ki} \\\\lambda_k$.\\nLet's look at $\\\\sum_{i=1}^{100} K_i \\\\cdot \\\\dots$?\\nConsider $K_j = \\\\sum_{i=1}^{100} a_{ij} \\\\lambda_i$.\\nMultiply by $x_j$?\\nWe have $\\\\sum_{j=1}^{100} x_j K_j = \\\\sum_{j=1}^{100} x_j \\\\sum_{i=1}^{100} a_{ij} \\\\lambda_i = \\\\sum_{i,j} a_{ij} x_j \\\\lambda_i$.\\nAlso consider rows. $R_i = \\\\sum_{j=1}^{100} a_{ij} \\\\lambda_j$.\\nIf we choose $x_j = 10^{k \\\\dots}$? No, $x_j$ should be related to $\\\\lambda_j$.\\nWhat if we take $\\\\sum_{j=1}^{100} R_j = \\\\sum_{j=1}^{100} \\\\sum_{i=1}^{100} a_{ij} \\\\lambda_j = \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{ij} \\\\lambda_j$.\\nAlso $\\\\sum_{i=1}^{100} K_i = \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{ij} \\\\lambda_i = \\\\sum_{j=1}^{100} \\\\sum_{i=1}^{100} a_{ij} \\\\lambda_i$.\\nThese are clearly equal. So $\\\\sum R_i = \\\\sum K_i$. This doesn't help much directly as both sums could be non-zero.\\n\\nLet's try to find weights $u_i$ for rows and $v_j$ for columns such that the double sum cancels out in a specific way.\\nOr maybe use the fact that $\\\\lambda_j = 10^{100-j}$.\\nRecall $R_i \\\\equiv 0 \\\\implies \\\\sum_j a_{ij} \\\\lambda_j \\\\equiv 0$.\\n$K_j \\\\equiv 0 \\\\implies \\\\sum_i a_{ij} \\\\lambda_i \\\\equiv 0$.\\nSumming $K_j$ over all $j$: $\\\\sum_j K_j = \\\\sum_j \\\\sum_i a_{ij} \\\\lambda_i = \\\\sum_i \\\\lambda_i (\\\\sum_j a_{ij})$.\\nSumming $R_i$ over all $i$: $\\\\sum_i R_i = \\\\sum_i \\\\sum_j a_{ij} \\\\lambda_j = \\\\sum_j \\\\lambda_j (\\\\sum_i a_{ij})$.\\nThis doesn't make them equal unless $a_{ij}$ is symmetric or something? No.\\nHowever, $\\\\sum_{i=1}^{100} K_i - \\\\sum_{j=1}^{100} R_j = \\\\sum_i \\\\lambda_i \\\\sum_j a_{ij} - \\\\sum_j \\\\lambda_j \\\\sum_i a_{ij}$.\\nLet $S_j = \\\\sum_i a_{ij}$ be the sum of column $j$ (digits).\\nThen $\\\\sum R_i = \\\\sum_{j=1}^{100} S_j \\\\lambda_j$? No.\\nWait, $R_i = \\\\sum_{j=1}^{100} a_{ij} \\\\lambda_j$.\\n$\\\\sum_{i=1}^{100} R_i = \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{ij} \\\\lambda_j = \\\\sum_{j=1}^{100} \\\\lambda_j \\\\left(\\\\sum_{i=1}^{100} a_{ij}\\\\right)$.\\nLet $S_{col, j} = \\\\sum_{i=1}^{100} a_{ij}$. Then $\\\\sum R_i = \\\\sum_{j=1}^{100} S_{col, j} \\\\lambda_j$.\\nAlso $K_j = \\\\sum_{i=1}^{100} a_{ij} \\\\lambda_i$.\\n$\\\\sum_{j=1}^{100} K_j = \\\\sum_{j=1}^{100} \\\\sum_{i=1}^{100} a_{ij} \\\\lambda_i = \\\\sum_{i=1}^{100} \\\\lambda_i \\\\left(\\\\sum_{j=1}^{100} a_{ij}\\\\right)$.\\nLet $S_{row, i} = \\\\sum_{j=1}^{100} a_{ij}$. Then $\\\\sum K_j = \\\\sum_{i=1}^{100} S_{row, i} \\\\lambda_i$.\\nNote that $S_{col, j} \\\\equiv K_j \\\\pmod 3$ ?\\n$K_j \\\\equiv \\\\sum a_{ij} \\\\pmod 3$. Since $\\\\lambda_i \\\\equiv 1 \\\\pmod 3$ (as $10 \\\\equiv 1 \\\\pmod 3$).\\nYes, $\\\\lambda_i = 10^{100-i} \\\\equiv 1^{100-i} \\\\equiv 1 \\\\pmod 3$.\\nSo $K_j \\\\equiv S_{row, j} \\\\pmod 3$? No.\\n$K_j = \\\\sum a_{ij} \\\\lambda_i$. Modulo 3, $\\\\lambda_i \\\\equiv 1$. So $K_j \\\\equiv \\\\sum a_{ij} \\\\pmod 3$.\\nWait, $S_{row, i} = \\\\sum_j a_{ij}$.\\nWait, $K_j$ sums over $i$. So it's sum of elements in column $j$.\\nLet $T_j = \\\\sum_{i=1}^{100} a_{ij}$. Then $K_j = \\\\sum_{i=1}^{100} a_{ij} \\\\lambda_i$.\\nModulo 3, $\\\\lambda_i \\\\equiv 1$, so $K_j \\\\equiv T_j \\\\pmod 3$.\\nModulo 9? $\\\\lambda_i = 10^{100-i}$. $10 \\\\equiv 1 \\\\pmod 9$. So $\\\\lambda_i \\\\equiv 1 \\\\pmod 9$.\\nActually $10^k \\\\equiv 1 \\\\pmod k$ is not true generally.\\nBut $10 \\\\equiv 1 \\\\pmod d$ for $d|9$.\\nSo $K_j \\\\equiv \\\\sum a_{ij} \\\\pmod 9$.\\nAlso $R_i = \\\\sum_j a_{ij} \\\\lambda_j$.\\nModulo 9, $\\\\lambda_j \\\\equiv 1$. So $R_i \\\\equiv \\\\sum_j a_{ij} \\\\pmod 9$.\\nLet's denote row sum $u_i = \\\\sum_j a_{ij}$ and column sum $v_j = \\\\sum_i a_{ij}$.\\nThen $R_i \\\\equiv u_i \\\\pmod 9$ and $K_j \\\\equiv v_j \\\\pmod 9$.\\nWait, $u_i = \\\\sum_j a_{ij}$ is the sum of digits in row $i$.\\n$v_j = \\\\sum_i a_{ij}$ is the sum of digits in column $j$.\\nThe condition is about $R_i$ and $K_j$ being divisible by 2013.\\nSo $R_i \\\\equiv 0 \\\\pmod{2013}$ implies $R_i \\\\equiv 0 \\\\pmod 9$ implies $u_i \\\\equiv 0 \\\\pmod 9$.\\nAlso $R_i \\\\equiv 0 \\\\pmod{11}$.\\n$R_i = \\\\sum_{j=1}^{100} a_{ij} 10^{100-j} \\\\pmod{11}$.\\nSince $10 \\\\equiv -1 \\\\pmod{11}$, $10^k \\\\equiv (-1)^k \\\\pmod{11}$.\\nSo $R_i \\\\equiv \\\\sum_{j=1}^{100} a_{ij} (-1)^{100-j} \\\\pmod{11}$.\\nLet $(-1)^{100-j} = (-1)^{-j} = (-1)^j$ since 100 is even.\\nSo $R_i \\\\equiv \\\\sum_{j=1}^{100} a_{ij} (-1)^j \\\\pmod{11}$.\\nSimilarly, $K_j = \\\\sum_{i=1}^{100} a_{ij} 10^{100-i} \\\\pmod{11}$.\\n$K_j \\\\equiv \\\\sum_{i=1}^{100} a_{ij} (-1)^{100-i} \\\\equiv \\\\sum_{i=1}^{100} a_{ij} (-1)^i \\\\pmod{11}$.\\n\\nThis looks like a sign-weighted sum.\\nLet $z_{ij} = a_{ij} (-1)^j$. Then $R_i \\\\equiv \\\\sum_{j=1}^{100} z_{ij} \\\\pmod{11}$.\\nLet $w_{ij} = a_{ij} (-1)^i$. Then $K_j \\\\equiv \\\\sum_{i=1}^{100} w_{ij} \\\\pmod{11}$.\\nNote $z_{ij}$ and $w_{ij}$ differ by sign depending on parity of $i+j$.\\n$w_{ij} = a_{ij} (-1)^i = a_{ij} (-1)^j (-1)^{i-j}$. If $i-j$ is even, $w_{ij} = z_{ij}$. If odd, $w_{ij} = -z_{ij}$.\\nWait, simpler:\\n$R_i \\\\equiv \\\\sum_j a_{ij} (-1)^j \\\\pmod{11}$.\\n$K_j \\\\equiv \\\\sum_i a_{ij} (-1)^i \\\\pmod{11}$.\\nThis implies that $\\\\sum_i R_i$?\\nLet's try to sum $\\\\sum R_i$ mod 11.\\n$\\\\sum R_i \\\\equiv \\\\sum_{i,j} a_{ij} (-1)^j \\\\pmod{11} = \\\\sum_j (-1)^j (\\\\sum_i a_{ij}) = \\\\sum_j v_j (-1)^j$.\\nSimilarly $\\\\sum K_j \\\\equiv \\\\sum_{i,j} a_{ij} (-1)^i \\\\pmod{11} = \\\\sum_i (-1)^i (\\\\sum_j a_{ij}) = \\\\sum_i u_i (-1)^i$.\\nWait, $v_j$ is the sum of column $j$, $u_i$ is the sum of row $i$.\\nThis doesn't seem to give immediate contradictions.\\nBut we have the assumption that almost all are 0.\\nThere are 200 numbers. Suppose $R_1, \\\\dots, R_{100}$ are $0 \\\\pmod{11}$ except maybe $R_k$.\\nAnd $K_1, \\\\dots, K_{100}$ are $0 \\\\pmod{11}$ except maybe $K_l$.\\nIf all are $0 \\\\pmod{11}$, then $u_i \\\\equiv 0$ is not required, only $\\\\sum a_{ij} (-1)^j = 0$.\\nHowever, we established earlier $R_i \\\\equiv u_i \\\\pmod 9$.\\nSo if all $R_i, K_j$ are divisible by 2013, they are divisible by 9.\\nThus $u_i \\\\equiv 0 \\\\pmod 9$ for all $i$, and $v_j \\\\equiv 0 \\\\pmod 9$ for all $j$.\\nThis means the sum of digits in every row is divisible by 9, and every column.\\nIs it possible that $u_i \\\\not\\\\equiv 0 \\\\pmod 9$?\\nNo, because if $R_i \\\\equiv 0 \\\\pmod{2013}$, then $R_i \\\\equiv 0 \\\\pmod 9$.\\nSo if 100 rows are divisible, their digit sums are $0 \\\\pmod 9$.\\nSame for columns.\\nSuppose exactly one row $R_1$ is NOT divisible by 2013.\\nThen $R_1 \\\\not\\\\equiv 0 \\\\pmod{2013}$.\\nThis implies $R_1 \\\\not\\\\equiv 0 \\\\pmod 9$ OR $R_1 \\\\not\\\\equiv 0 \\\\pmod{11}$ (or 61).\\nCase 1: $R_1 \\\\not\\\\equiv 0 \\\\pmod 9$. Then $u_1 \\\\not\\\\equiv 0 \\\\pmod 9$.\\nBut columns $K_1, \\\\dots, K_{100}$ are ALL divisible by 2013?\\nWait, if $R_1$ fails, can it be that $K_1$ also fails?\\nWe are assuming at most 1 failure.\\nSo if $R_1$ fails, then $K_1, \\\\dots, K_{100}$ must be divisible by 2013.\\nIf $K_j$ are all divisible by 2013, then $v_j \\\\equiv 0 \\\\pmod 9$ for all $j=1,\\\\dots,100$.\\nSo $\\\\sum_i a_{ij} \\\\equiv 0 \\\\pmod 9$.\\nSumming over $j$, total sum $S = \\\\sum_{i,j} a_{ij} \\\\equiv 0 \\\\pmod 9$.\\nOn the other hand, $u_1 = \\\\sum_j a_{1j}$ (row 1 sum).\\nAlso $u_2 = \\\\sum_j a_{2j}, \\\\dots$.\\nSince $R_2, \\\\dots, R_{100}$ are divisible by 2013, $u_2, \\\\dots, u_{100} \\\\equiv 0 \\\\pmod 9$.\\nSo $S = u_1 + \\\\sum_{i=2}^{100} u_i \\\\equiv u_1 \\\\pmod 9$.\\nSo $u_1 \\\\equiv S \\\\equiv 0 \\\\pmod 9$.\\nBut $R_1$ not divisible by 2013 implies $R_1 \\\\not\\\\equiv 0 \\\\pmod 9$ is a possibility.\\nWait, does $R_1 \\\\not\\\\equiv 0 \\\\pmod{2013}$ imply $R_1 \\\\not\\\\equiv 0 \\\\pmod 9$?\\nNot necessarily. It could be that $R_1 \\\\equiv 0 \\\\pmod 9$ but not $0 \\\\pmod{11}$ (or 61).\\nSo we cannot immediately derive a contradiction from mod 9 alone.\\nHowever, if $R_1$ fails, then $K_1, \\\\dots, K_{100}$ succeed.\\n$K_j \\\\equiv 0 \\\\pmod{2013} \\\\implies K_j \\\\equiv 0 \\\\pmod{11}$.\\n$K_j \\\\equiv \\\\sum_i a_{ij} (-1)^i \\\\pmod{11} \\\\equiv 0$.\\nSo $\\\\sum_i a_{ij} (-1)^i \\\\equiv 0 \\\\pmod{11}$ for all $j$.\\nAlso $R_2, \\\\dots, R_{100} \\\\equiv 0 \\\\pmod{2013} \\\\implies R_k \\\\equiv 0 \\\\pmod{11}$.\\n$R_k \\\\equiv \\\\sum_j a_{kj} (-1)^j \\\\equiv 0 \\\\pmod{11}$.\\nWhat about $R_1$? $R_1 \\\\not\\\\equiv 0 \\\\pmod{2013}$.\\nThis might mean $R_1 \\\\not\\\\equiv 0 \\\\pmod{11}$. Or maybe $R_1 \\\\equiv 0 \\\\pmod{11}$ but fail elsewhere.\\nLet's analyze the condition $K_j \\\\equiv 0 \\\\pmod{11}$ and $R_k \\\\equiv 0 \\\\pmod{11}$.\\nWe have a matrix $A = (a_{ij})$.\\nCondition $C_j$: $\\\\sum_{i=1}^{100} a_{ij} (-1)^i \\\\equiv 0 \\\\pmod{11}$.\\nCondition $R_k$: $\\\\sum_{j=1}^{100} a_{kj} (-1)^j \\\\equiv 0 \\\\pmod{11}$.\\nLet $b_{ij} = a_{ij} (-1)^i$. Then $C_j \\\\equiv \\\\sum_i b_{ij} \\\\equiv 0 \\\\pmod{11}$.\\nLet $d_{kj} = a_{kj} (-1)^j$. Then $R_k \\\\equiv \\\\sum_j d_{kj} \\\\equiv 0 \\\\pmod{11}$.\\nIf we assume $R_1$ is the only failure. Then $R_1 \\\\not\\\\equiv 0 \\\\pmod{11}$? Not necessarily, but if $R_1 \\\\equiv 0 \\\\pmod{11}$, it still could fail mod 61.\\nLet's focus on 11 first.\\nAssume $K_1, \\\\dots, K_{100} \\\\equiv 0 \\\\pmod{11}$ and $R_2, \\\\dots, R_{100} \\\\equiv 0 \\\\pmod{11}$.\\nFrom $K_j \\\\equiv 0 \\\\implies \\\\sum_{i} a_{ij} (-1)^i \\\\equiv 0$.\\nThis means the alternating column sum is 0.\\nFrom $R_k \\\\equiv 0 (k \\\\ge 2) \\\\implies \\\\sum_{j} a_{kj} (-1)^j \\\\equiv 0$.\\nThis means row alternating sum is 0.\\nLet's check the consistency.\\n$\\\\sum_{j=1}^{100} \\\\sum_{i=1}^{100} a_{ij} (-1)^i (-1)^j$ ?\\nConsider $\\\\sum_j (-1)^j K_j = \\\\sum_j (-1)^j \\\\sum_i a_{ij} (-1)^i = \\\\sum_{i,j} a_{ij} (-1)^{i+j}$.\\nConsider $\\\\sum_i (-1)^i R_i = \\\\sum_i (-1)^i \\\\sum_j a_{ij} (-1)^j = \\\\sum_{i,j} a_{ij} (-1)^{i+j}$.\\nThey are equal.\\nSo $\\\\sum_j (-1)^j K_j = \\\\sum_i (-1)^i R_i$.\\nModulo 11.\\nIf $K_j \\\\equiv 0$ for all $j$, LHS is 0.\\nIf $R_i \\\\equiv 0$ for all $i$, RHS is 0. Consistent.\\nIf $K_j \\\\equiv 0$ for all $j$, and $R_1 \\\\not\\\\equiv 0$.\\nRHS $= (-1)^1 R_1 + \\\\sum_{i=2}^{100} (-1)^i (0) = -R_1$.\\nSo $-R_1 \\\\equiv 0 \\\\pmod{11}$, which implies $R_1 \\\\equiv 0 \\\\pmod{11}$.\\nSo even if $R_1$ fails to be divisible by 2013, it MUST be divisible by 11.\\nWait, if $R_1$ is not divisible by 2013, could it fail only due to mod 61?\\nYes.\\nSo we know $R_1 \\\\equiv 0 \\\\pmod{11}$.\\nDoes this force anything else?\\nWe have $K_j \\\\equiv 0 \\\\pmod{11}$ for all $j$.\\n$R_1 \\\\equiv 0 \\\\pmod{11}$ is derived from $\\\\sum (-1)^j K_j = \\\\sum (-1)^i R_i$.\\nWait, I assumed $K_j$ all zero.\\nWhat if $K_1$ is the failure?\\nIf $R_1, \\\\dots, R_{100}$ all divisible, $K_1$ not divisible.\\nThen $\\\\sum K_j = \\\\sum_i R_i$. No, $\\\\sum (-1)^j K_j = \\\\sum (-1)^i R_i$.\\nIf $R_i$ all 0, RHS = 0.\\nLHS $= (-1)^1 K_1 + \\\\sum_{j=2}^{100} 0 = -K_1$.\\nSo $K_1 \\\\equiv 0 \\\\pmod{11}$.\\nThus, if all 100 rows are divisible by 11, then ALL columns must be divisible by 11.\\nConversely, if all 100 columns are divisible by 11, all rows must be divisible by 11.\\nProof: $\\\\sum_{j} (-1)^j K_j = \\\\sum_{i} (-1)^i R_i$.\\nLet $S_K = \\\\sum (-1)^j K_j$ and $S_R = \\\\sum (-1)^i R_i$.\\nIf all $K_j \\\\equiv 0 \\\\pmod{11}$, $S_K \\\\equiv 0 \\\\implies S_R \\\\equiv 0$.\\n$S_R = R_1 - R_2 + R_3 - \\\\dots + R_{100}$.\\nIf $R_2, \\\\dots, R_{100}$ are divisible by 11, then $S_R \\\\equiv R_1 \\\\pmod{11}$.\\nSo $R_1 \\\\equiv 0 \\\\pmod{11}$.\\nWait, the assumption was $K_j$ all 0. $R_i$ are arbitrary?\\nIf $K_j$ all 0, then $S_R=0$.\\nIf $R_2, \\\\dots, R_{100} \\\\equiv 0$, then $R_1 \\\\equiv 0$.\\nIf $R_1, \\\\dots, R_{99} \\\\equiv 0$, what about $R_{100}$?\\n$S_R = \\\\sum_{i=1}^{99} 0 + (-1)^{100} R_{100} = R_{100}$. So $R_{100} \\\\equiv 0$.\\nBasically, if 100 rows are divisible, 100 cols are divisible.\\nWhat if 1 row is missing?\\nSuppose $R_1 \\\\not\\\\equiv 0 \\\\pmod{11}$. Then we must have some $K_j \\\\not\\\\equiv 0 \\\\pmod{11}$ to balance?\\nFrom $\\\\sum (-1)^j K_j = \\\\sum (-1)^i R_i$.\\nSuppose we have at most 1 failure in total.\\nPossible scenarios for mod 11 failures:\\nLet $n_r$ be number of rows not div by 11. $n_c$ be number of cols not div by 11.\\nWe know $n_r + n_c \\\\le 1$.\\nSo either $(n_r, n_c) = (0,0)$, $(1,0)$, or $(0,1)$.\\nCase A: $(0,0)$. All divisible by 11. OK.\\nCase B: $(1,0)$. One row $R_k \\\\not\\\\equiv 0$, all $C_j \\\\equiv 0$.\\nEquation: $\\\\sum (-1)^j C_j = \\\\sum (-1)^i R_i \\\\implies 0 = \\\\sum (-1)^i R_i$.\\nSince $C_j \\\\equiv 0$ for all $j$, sum is 0.\\nSo $\\\\sum (-1)^i R_i \\\\equiv 0$.\\nSince only $R_k$ is non-zero, $\\\\sum (-1)^i R_i = (-1)^k R_k$.\\nSo $(-1)^k R_k \\\\equiv 0 \\\\implies R_k \\\\equiv 0$. Contradiction.\\nCase C: $(0,1)$. All $R_i \\\\equiv 0$, one col $C_k \\\\not\\\\equiv 0$.\\nEquation: $\\\\sum (-1)^j C_j = \\\\sum (-1)^i R_i \\\\implies \\\\sum (-1)^j C_j = 0$.\\nOnly $C_k$ is non-zero, so $(-1)^k C_k \\\\equiv 0 \\\\implies C_k \\\\equiv 0$. Contradiction.\\n\\nConclusion: It is impossible to have exactly 1 failure mod 11.\\nEither all 200 numbers are divisible by 11, or it's impossible to satisfy the \\\"at most 1 failure\\\" constraint for 11?\\nWait. This logic holds IF the equation $\\\\sum (-1)^j C_j = \\\\sum (-1)^i R_i$ is correct.\\nLet's verify.\\n$C_j = \\\\sum_i a_{ij} 10^{100-i}$.\\nMod 11: $10 \\\\equiv -1$. $10^{100-i} \\\\equiv (-1)^{100-i} = (-1)^{-i} = (-1)^i$ (since 100 is even).\\nSo $C_j \\\\equiv \\\\sum_i a_{ij} (-1)^i$.\\nThen $\\\\sum_j (-1)^j C_j \\\\equiv \\\\sum_j (-1)^j \\\\sum_i a_{ij} (-1)^i = \\\\sum_{i,j} a_{ij} (-1)^{i+j}$.\\n$R_i = \\\\sum_j a_{ij} 10^{100-j}$.\\nMod 11: $R_i \\\\equiv \\\\sum_j a_{ij} (-1)^j$.\\nThen $\\\\sum_i (-1)^i R_i \\\\equiv \\\\sum_i (-1)^i \\\\sum_j a_{ij} (-1)^j = \\\\sum_{i,j} a_{ij} (-1)^{i+j}$.\\nSo the equality holds.\\nTherefore, $\\\\sum (-1)^j C_j \\\\equiv \\\\sum (-1)^i R_i \\\\pmod{11}$.\\nLet $E_R = \\\\sum_{i=1}^{100} (-1)^i R_i$ and $E_C = \\\\sum_{j=1}^{100} (-1)^j C_j$.\\n$E_R \\\\equiv E_C \\\\pmod{11}$.\\nIf we have at most 1 failure.\\nIf there are 0 failures, $0 \\\\equiv 0$.\\nIf there is 1 failure, say $R_k$. Then $E_R \\\\equiv (-1)^k R_k \\\\not\\\\equiv 0$ (if $R_k$ not div by 11).\\nWait, $R_k$ is not div by 2013. Does it mean $R_k \\\\not\\\\equiv 0 \\\\pmod{11}$?\\nNot necessarily. $R_k$ could be $0 \\\\pmod{11}$ but not $0 \\\\pmod{2013}$.\\nExample: $R_k = 11 \\\\times 61 = 671$. $R_k \\\\equiv 0 \\\\pmod{11}$, $R_k \\\\not\\\\equiv 0 \\\\pmod{2013}$.\\nSo the failure could be hidden in 61.\\nHowever, we are asked to prove ALL are divisible by 2013.\\nSo we must prove that if there is a failure in 2013, we reach a contradiction? Or prove that failure leads to contradiction.\\nWait, if $R_k \\\\equiv 0 \\\\pmod{11}$ but $R_k \\\\not\\\\equiv 0 \\\\pmod{2013}$, then the modulo 11 condition gives $0 \\\\equiv 0$, so no contradiction yet.\\nHowever, we need to handle mod 61 or mod 3.\\nAlso mod 3.\\nRecall $R_i \\\\equiv \\\\sum a_{ij} \\\\pmod 9$.\\n$C_j \\\\equiv \\\\sum a_{ij} \\\\pmod 9$. (Using $10 \\\\equiv 1 \\\\pmod 9$).\\nLet $U_i = R_i \\\\pmod 9$ and $V_j = C_j \\\\pmod 9$.\\n$U_i \\\\equiv \\\\sum_j a_{ij} \\\\pmod 9$.\\n$V_j \\\\equiv \\\\sum_i a_{ij} \\\\pmod 9$.\\nWait, earlier I said $C_j \\\\equiv \\\\sum_i a_{ij} \\\\pmod 9$.\\nCheck: $C_j = \\\\sum_i a_{ij} 10^{100-i}$.\\nSince $10 \\\\equiv 1 \\\\pmod 9$, $10^{100-i} \\\\equiv 1$. So $C_j \\\\equiv \\\\sum_i a_{ij} \\\\pmod 9$. Correct.\\nAnd $R_i \\\\equiv \\\\sum_j a_{ij} \\\\pmod 9$. Correct.\\nAlso note $\\\\sum_i U_i = \\\\sum_i \\\\sum_j a_{ij} = \\\\sum_{i,j} a_{ij}$.\\nAnd $\\\\sum_j V_j = \\\\sum_j \\\\sum_i a_{ij} = \\\\sum_{i,j} a_{ij}$.\\nSo $\\\\sum U_i = \\\\sum V_j$.\\nLet $S = \\\\sum_{i,j} a_{ij}$.\\nIf we have at most 1 failure for 2013.\\nDoes this imply at most 1 failure for mod 9?\\nFailure of 2013 $\\\\implies$ failure of at least one of 3, 11, 61.\\nDivisibility by 3 is divisibility by 9? No.\\nBut if $X \\\\equiv 0 \\\\pmod{2013}$, then $X \\\\equiv 0 \\\\pmod 3$.\\nIf $X \\\\not\\\\equiv 0 \\\\pmod{2013}$, then $X$ might be $0 \\\\pmod 3$.\\nSo having 199 divisible by 2013 implies 199 divisible by 3.\\nActually, $2013 = 3 \\\\times 11 \\\\times 61$.\\nIf $N$ is divisible by 2013, it is divisible by 3.\\nIf we have 199 numbers divisible by 2013, do we have restrictions on mod 3?\\nSuppose 199 are divisible by 2013, so 199 divisible by 3.\\nThe last one might not be divisible by 3.\\nOr it might be.\\nIf it is not divisible by 3, then $u_i \\\\not\\\\equiv 0 \\\\pmod 3$ or $v_j \\\\not\\\\equiv 0 \\\\pmod 3$.\\nLet's refine this.\\nSuppose there is exactly 1 failure in mod 2013.\\nIf the failure is at $R_k$, then $R_k$ is not divisible by 2013.\\nThis implies $R_k \\\\not\\\\equiv 0 \\\\pmod 3$ OR $R_k \\\\not\\\\equiv 0 \\\\pmod{11}$ OR $R_k \\\\not\\\\equiv 0 \\\\pmod{61}$.\\nHowever, we proved earlier that for mod 11, if all cols are div, all rows must be div.\\nWait, did we?\\nWe showed $E_R = E_C$.\\nIf $C_j \\\\equiv 0$ for all $j$, then $E_C \\\\equiv 0$. Thus $E_R \\\\equiv 0$.\\n$E_R = \\\\sum (-1)^i R_i$.\\nIf $R_i \\\\equiv 0$ for all $i \\\\neq k$, then $E_R \\\\equiv (-1)^k R_k$.\\nSo if all $C_j \\\\equiv 0$, then $R_k \\\\equiv 0 \\\\pmod{11}$.\\nThis means: If all columns are divisible by 11, then the potential row failure MUST be divisible by 11.\\nSimilarly, if all rows are divisible by 11, the potential column failure MUST be divisible by 11.\\nThis restricts the location of the failure.\\nIf $R_k$ is not divisible by 11, then $E_R \\\\not\\\\equiv 0$ (assuming others are 0 mod 11).\\nThis would require $E_C \\\\not\\\\equiv 0$, so at least one $C_j$ is not divisible by 11.\\nBut we only allow 1 failure total. So if $R_k$ fails mod 11, then $C_1, \\\\dots, C_{100}$ must all be divisible by 11.\\nBut $E_R = (-1)^k R_k \\\\not\\\\equiv 0$. $E_C = 0$. Contradiction.\\nSo: You cannot have exactly one failure which is divisible by 2013 failing ONLY because of mod 11 divisibility (or rather, not divisible by 11).\\nMore precisely:\\nIt is impossible to have a configuration where exactly one number $X \\\\in \\\\{R_i, C_j\\\\}$ satisfies $X \\\\not\\\\equiv 0 \\\\pmod{11}$ while all others are $0 \\\\pmod{11}$.\\nWait, the condition is \\\"at least 199 divisible by 2013\\\".\\nThis means at most 1 is NOT divisible by 2013.\\nIt does NOT say at most 1 is not divisible by 11.\\nBecause $X$ could be divisible by 11 but not 3 or 61.\\nIf $X$ is not divisible by 2013, then $X \\\\not\\\\equiv 0 \\\\pmod{2013}$.\\nThis allows $X \\\\equiv 0 \\\\pmod{11}$ but $X \\\\not\\\\equiv 0 \\\\pmod 3$.\\nIn this case, $X \\\\equiv 0 \\\\pmod{11}$ contributes 0 to $E_R$ or $E_C$ mod 11.\\nSo mod 11 analysis doesn't rule out the existence of a failure.\\nIt just says if the failure is a multiple of 11, then $E_R=0$ and $E_C=0$ are satisfied regardless.\\nWait, if $R_k$ fails, it means $R_k \\\\not\\\\equiv 0 \\\\pmod{2013}$.\\nIf $R_k \\\\equiv 0 \\\\pmod{11}$, then it contributes 0 to $E_R$.\\nBut $R_k$ might be non-zero.\\nLet's check the constraint again.\\n\\\"At least 199 of these 200 numbers are divisible by 2013.\\\"\\nSuppose $R_1$ is the unique non-divisible number.\\nThen $R_1 \\\\not\\\\equiv 0 \\\\pmod{2013}$.\\nAll $R_i$ for $i>1$ are divisible by 2013 $\\\\implies$ divisible by 11.\\nAll $C_j$ for all $j$ are divisible by 2013 $\\\\implies$ divisible by 11.\\nSo we have a setup where all $R_i (i>1)$ are $0 \\\\pmod{11}$ and all $C_j$ are $0 \\\\pmod{11}$.\\nAs shown before, this implies $R_1 \\\\equiv 0 \\\\pmod{11}$.\\nSo $R_1$ is divisible by 11.\\nThus $R_1$ fails to be divisible by 2013 ONLY because of mod 3 or mod 61.\\n(Since $2013 = 3 \\\\cdot 11 \\\\cdot 61$).\\nThis reduces the problem:\\nWe have $R_1 \\\\equiv 0 \\\\pmod{11}$ (and $C_j \\\\equiv 0 \\\\pmod{11}$, etc if we pick another failure scenario).\\nAlso we have at most 1 failure.\\nNow consider Modulo 3.\\n$R_i \\\\equiv \\\\sum_j a_{ij} \\\\pmod 3$.\\n$C_j \\\\equiv \\\\sum_i a_{ij} \\\\pmod 3$.\\nIf $R_1$ is the failure, then $R_1$ might or might not be divisible by 3.\\nIf $R_1$ is divisible by 3, then it works mod 3.\\nIf it is not divisible by 3, then $\\\\sum a_{1j} \\\\not\\\\equiv 0 \\\\pmod 3$.\\nAlso all other rows $R_2, \\\\dots, R_{100}$ are divisible by 3.\\nAll columns $C_1, \\\\dots, C_{100}$ are divisible by 3.\\nWait. If $C_j$ are all divisible by 3, then $\\\\sum_i a_{ij} \\\\equiv 0 \\\\pmod 3$ for all $j$.\\nThen $\\\\sum_{i,j} a_{ij} = \\\\sum_j (\\\\sum_i a_{ij}) \\\\equiv 0 \\\\pmod 3$.\\nAlso $\\\\sum_i R_i = \\\\sum_i (\\\\sum_j a_{ij}) \\\\equiv 0 \\\\pmod 3$?\\n$\\\\sum_i R_i \\\\equiv \\\\sum_j (\\\\sum_i a_{ij}) \\\\equiv \\\\sum_j 0 \\\\equiv 0 \\\\pmod 3$.\\nIf $R_1$ is not divisible by 3, then $\\\\sum_{i=2}^{100} R_i + R_1 \\\\not\\\\equiv 0 \\\\pmod 3$?\\nWait. $\\\\sum_i R_i \\\\equiv 0 \\\\pmod 3$ was derived from $\\\\sum_i R_i \\\\equiv \\\\sum_{i,j} a_{ij}$.\\nWait, $\\\\sum_{i,j} a_{ij} = \\\\sum_j (\\\\sum_i a_{ij})$.\\nIf all $C_j$ are divisible by 3, then $\\\\sum_i a_{ij} \\\\equiv 0 \\\\pmod 3$.\\nSo sum of digits is $\\\\equiv 0 \\\\pmod 3$.\\nThen $\\\\sum_i R_i \\\\equiv \\\\sum_j (\\\\sum_i a_{ij}) \\\\equiv \\\\sum_j 0 \\\\equiv 0 \\\\pmod 3$?\\nNO.\\n$R_i = \\\\sum_j a_{ij} 10^{100-j}$.\\n$R_i \\\\equiv \\\\sum_j a_{ij} \\\\pmod 3$.\\nSo $\\\\sum_i R_i \\\\equiv \\\\sum_i \\\\sum_j a_{ij} = \\\\sum_{i,j} a_{ij}$.\\nThis equals $\\\\sum_j \\\\sum_i a_{ij}$.\\nIf all $C_j$ are divisible by 3, does that mean $\\\\sum_i a_{ij} \\\\equiv 0 \\\\pmod 3$?\\nYes, because $C_j \\\\equiv \\\\sum_i a_{ij} \\\\pmod 3$.\\nSo if all $C_j$ are divisible by 3, then $\\\\sum_{i,j} a_{ij} \\\\equiv 0 \\\\pmod 3$.\\nThis implies $\\\\sum_i R_i \\\\equiv 0 \\\\pmod 3$.\\nSo $\\\\sum_{i=1}^{100} R_i \\\\equiv 0 \\\\pmod 3$.\\nIf $R_1$ is the unique failure, then $R_2, \\\\dots, R_{100}$ are divisible by 3.\\nSo $R_1 \\\\equiv -\\\\sum_{i=2}^{100} R_i \\\\equiv 0 \\\\pmod 3$.\\nSo $R_1$ MUST be divisible by 3.\\nSimilar logic applies to columns.\\nWait. In our scenario, we assumed $R_1$ is the unique failure.\\nSo $C_j$ are all divisible by 3.\\nThis forces $R_1$ to be divisible by 3.\\nThus, the failure cannot be due to mod 3.\\nSo the failure must be due to mod 61.\\nWait, this is promising.\\nLet's formalize.\\nHypothesis: Exactly one number, say $X$, is not divisible by 2013. All others are.\\nStep 1: Analyze Modulo 11.\\nLet $N$ be the set of indices for rows/cols. $|N|=200$.\\nDefine $f(R_i) = R_i \\\\pmod{11}$, $g(C_j) = C_j \\\\pmod{11}$.\\nLet $A_{11} = \\\\{ X \\\\in \\\\text{Rows} \\\\cup \\\\text{Cols} : X \\\\not\\\\equiv 0 \\\\pmod{11} \\\\}$.\\nWe know $|A_{2013}| \\\\le 1$.\\nWe proved earlier:\\n$\\\\sum_{i} (-1)^i f(R_i) = \\\\sum_{j} (-1)^j g(C_j) \\\\pmod{11}$.\\nLet $S_R = \\\\sum (-1)^i f(R_i)$ and $S_C = \\\\sum (-1)^j g(C_j)$.\\nIf $X \\\\notin A_{11}$ (i.e. divisible by 11), it contributes 0 to the sum.\\nIf $A_{11}$ contains an element, the sum is non-zero.\\nSpecifically, if exactly one element $X$ is not divisible by 2013, it must contribute to the sums.\\nWait, if $X$ is not divisible by 2013, it *could* still be divisible by 11.\\nIf $X$ is divisible by 11, then $S_R = 0$ and $S_C = 0$.\\nIf $X$ is NOT divisible by 11, then $S_R \\\\neq 0$ or $S_C \\\\neq 0$ or both.\\nWait, if $X$ is a row $R_k$, then $S_R = (-1)^k R_k \\\\pmod{11}$.\\nIf $R_k \\\\not\\\\equiv 0 \\\\pmod{11}$, $S_R \\\\neq 0$.\\nThen we need $S_C \\\\neq 0$ to satisfy $S_R = S_C$.\\nBut $S_C = \\\\sum (-1)^j C_j$. If $C_j$ are all $0$ (since only $R_k$ fails and we assume only 1 failure in total, wait... NO).\\nThe assumption is \\\"At most 1 failure of 2013\\\".\\nIt does NOT say \\\"At most 1 failure of 11\\\".\\nBecause if $R_k$ fails 2013 but passes 11, then all $C_j$ pass 11 (maybe) or some fail 11?\\nWait, if $R_k$ fails 2013, and passes 11. Then $R_k \\\\equiv 0 \\\\pmod{11}$.\\nThen $R_i \\\\equiv 0 \\\\pmod{11}$ for $i \\\\neq k$ (since they pass 2013).\\nAnd $C_j$ are either all passing 2013 (so $0 \\\\pmod{11}$) or maybe one fails 2013? No, only $R_k$ fails. So all $C_j$ pass 2013, so $C_j \\\\equiv 0 \\\\pmod{11}$.\\nIn this case ($R_k$ fails 2013, passes 11), we have $f(R_i)=0$ for all $i$, $g(C_j)=0$ for all $j$.\\nThen $S_R = 0, S_C = 0$. Equation holds. No contradiction.\\nSo we cannot deduce $R_k \\\\not\\\\equiv 0 \\\\pmod{11}$.\\nWait, I previously argued \\\"impossible to have exactly 1 failure mod 11\\\".\\nLet's re-evaluate.\\nThe failure is defined by divisibility by 2013.\\nWe are trying to prove: IF at least 199 are div by 2013, THEN all are.\\nContrapositive: If not all are, then at least 2 are not div by 2013? Or rather, if at most 198 are div by 2013, then ... No.\\nDirect proof: Assume at most 1 is not div by 2013. Let this be set $F$. $|F| \\\\le 1$.\\nIf $F = \\\\emptyset$, done.\\nAssume $|F|=1$. Let $X \\\\in F$.\\nWe want to derive a contradiction.\\nWe found that $S_R = S_C \\\\pmod{11}$.\\nWe know that for any $Y \\\\notin F$, $Y$ is divisible by 2013 $\\\\implies Y \\\\equiv 0 \\\\pmod{11}$.\\nFor $X \\\\in F$, $X$ might be $0 \\\\pmod{11}$ or not.\\nIf $X \\\\not\\\\equiv 0 \\\\pmod{11}$, then $X$ is the sole element contributing to the sums.\\nIf $X$ is a row $R_k$, $S_R = \\\\pm R_k \\\\not\\\\equiv 0$. $S_C = 0$ (since all $C_j$ are $0 \\\\pmod{11}$).\\nContradiction! So $X$ cannot be a row that is not divisible by 11.\\nSimilarly, if $X$ is a column not divisible by 11, $S_C \\\\neq 0, S_R = 0$, contradiction.\\nTherefore, if there is exactly one failure $X$, it MUST satisfy $X \\\\equiv 0 \\\\pmod{11}$.\\n(This is a key step).\\n\\nSo, $X \\\\equiv 0 \\\\pmod{11}$.\\nNow consider modulo 3.\\nLet $h(Y) = Y \\\\pmod 3$.\\nWe have $\\\\sum h(R_i) \\\\equiv \\\\sum h(C_j) \\\\pmod 3$?\\nLet's check.\\n$h(R_i) \\\\equiv \\\\sum_j a_{ij} \\\\pmod 3$.\\n$h(C_j) \\\\equiv \\\\sum_i a_{ij} \\\\pmod 3$.\\n$\\\\sum_i h(R_i) \\\\equiv \\\\sum_{i,j} a_{ij}$.\\n$\\\\sum_j h(C_j) \\\\equiv \\\\sum_{j,i} a_{ij}$.\\nYes, $\\\\sum h(R_i) = \\\\sum h(C_j) \\\\pmod 3$.\\nLet $S'_R = \\\\sum_i h(R_i)$ and $S'_C = \\\\sum_j h(C_j)$.\\n$S'_R = S'_C$.\\nAgain, all $Y \\\\notin F$ are divisible by 3.\\nIf $X$ is the unique failure, then all $Y \\\\in \\\\{Others\\\\}$ are $0 \\\\pmod 3$.\\nSo $S'_R$ depends only on $X$ if $X$ is a row.\\nIf $X = R_k$, then $S'_R = h(R_k)$, $S'_C = \\\\sum_j 0 = 0$.\\nSo $h(R_k) = 0$.\\nThus $R_k \\\\equiv 0 \\\\pmod 3$.\\nIf $X = C_k$, then $S'_C = h(C_k) \\\\neq 0 \\\\implies$ contradiction with $S'_R = 0$.\\nSo $h(C_k) = 0$.\\nConclusion: The unique failure $X$ MUST be divisible by 3.\\nCombining with previous result: $X \\\\equiv 0 \\\\pmod 3$ AND $X \\\\equiv 0 \\\\pmod{11}$.\\nSince 3 and 11 are coprime, $X \\\\equiv 0 \\\\pmod{33}$.\\nSo $X$ is divisible by 33.\\nBut $X$ is NOT divisible by 2013.\\nThus, $X$ must be divisible by 61?\\nNo, $X$ is NOT divisible by 2013.\\n$2013 = 33 \\\\times 61$.\\nSince $X$ is div by 33, for $X$ to NOT be div by 2013, $X$ must NOT be divisible by 61.\\nSo $X$ is divisible by 33 but not 61.\\nAlso, all other numbers are divisible by 2013, hence divisible by 61.\\nWait.\\nThis logic seems to lead nowhere yet. It just says if there is 1 failure, it must be $0 \\\\pmod{33}$ but $\\\\not\\\\equiv 0 \\\\pmod{61}$.\\nThis is possible. e.g. $X = 33 \\\\times 2 = 66$.\\nBut we need to show this situation is impossible.\\nLet's analyze modulo 61.\\nWe need a linear relation between $R_i$ and $C_j$ mod 61.\\nUnlike mod 3 and 11, where $10 \\\\equiv 1$ and $10 \\\\equiv -1$.\\nFor mod 61, $10$ is a primitive root? Or something else.\\n$10^k \\\\pmod{61}$.\\n$10^1 = 10$.\\n$10^2 = 100 \\\\equiv 39 \\\\equiv -22$.\\n$10^3 \\\\equiv -220 = -220 + 3\\\\times 61 = -220 + 183 = -37 \\\\equiv 24$.\\n$10^4 \\\\equiv 240 = 4 \\\\times 61 - 4$. $4 \\\\times 61 = 244$. So $240 \\\\equiv -4$.\\n$10^5 \\\\equiv -40 \\\\equiv 21$.\\n...\\nThis seems tedious to calculate manually. However, there is a known trick for these grid divisibility problems.\\nOften the trick is to combine equations with weights.\\nRecall $S_R = S_C$ mod $p$.\\nFor general $m$, can we define $S_m = \\\\sum c_i R_i - \\\\sum d_j C_j \\\\equiv 0$?\\nWe had for $m=11$, weights $(-1)^i$.\\nFor $m=3$, weights $1$.\\nFor $m=61$, maybe weights $10^{k}$?\\nLet's go back to $R_i = \\\\sum_j a_{ij} \\\\lambda_j$ and $C_j = \\\\sum_i a_{ij} \\\\lambda_i$.\\nwhere $\\\\lambda_k = 10^{100-k}$.\\nConsider the sum $\\\\sum_{i=1}^{100} R_i$?\\nNo, we want a relation independent of $a_{ij}$.\\nLook at the relation $\\\\sum_{j=1}^{100} K_j \\\\lambda_j^{-1}$? No.\\nLet's try to eliminate $a_{ij}$.\\nConsider the polynomial $P(x) = \\\\sum_{j=1}^{100} a_{1j} x^{100-j}$.\\nWait, this doesn't capture the whole table.\\nLet's reconsider the identity $\\\\sum_{j=1}^{100} C_j (-1)^j \\\\equiv \\\\sum_{i=1}^{100} R_i (-1)^i \\\\pmod{11}$.\\nThis came from $\\\\lambda_j \\\\equiv (-1)^j$.\\nGenerally, $\\\\lambda_j = 10^{100-j}$.\\nWe want $\\\\sum_j K_j \\\\beta_j \\\\equiv \\\\sum_i R_i \\\\alpha_i \\\\pmod{m}$ for some coefficients $\\\\alpha, \\\\beta$.\\n$K_j = \\\\sum_i a_{ij} \\\\lambda_i$.\\n$\\\\sum_j K_j \\\\beta_j = \\\\sum_{i,j} a_{ij} \\\\lambda_i \\\\beta_j$.\\n$R_i = \\\\sum_j a_{ij} \\\\lambda_j$.\\n$\\\\sum_i R_i \\\\alpha_i = \\\\sum_{i,j} a_{ij} \\\\lambda_j \\\\alpha_i$.\\nTo match coefficients of $a_{ij}$, we need $\\\\lambda_i \\\\beta_j \\\\equiv \\\\lambda_j \\\\alpha_i \\\\pmod{m}$.\\nThis requires $\\\\frac{\\\\beta_j}{\\\\lambda_j} \\\\equiv \\\\frac{\\\\alpha_i}{\\\\lambda_i} \\\\pmod{m}$.\\nLet this common ratio be $\\\\gamma$.\\nSo $\\\\beta_j = \\\\gamma \\\\lambda_j$ and $\\\\alpha_i = \\\\gamma \\\\lambda_i$.\\nThen $\\\\sum_j K_j (\\\\gamma \\\\lambda_j) = \\\\sum_i R_i (\\\\gamma \\\\lambda_i)$.\\nThis simplifies to $\\\\gamma \\\\sum_j K_j \\\\lambda_j = \\\\gamma \\\\sum_i R_i \\\\lambda_i$.\\nIf $\\\\gamma \\\\not\\\\equiv 0$, we can divide.\\nLet's try $\\\\gamma = 1$.\\nIdentity: $\\\\sum_{j=1}^{100} K_j 10^{100-j} \\\\equiv \\\\sum_{i=1}^{100} R_i 10^{100-i} \\\\pmod m$.\\nWait.\\nLHS: $\\\\sum_{j=1}^{100} (\\\\sum_{i=1}^{100} a_{ij} 10^{100-i}) 10^{100-j} = \\\\sum_{i,j} a_{ij} 10^{100-i} 10^{100-j} = \\\\sum_{i,j} a_{ij} 10^{200-i-j}$.\\nRHS: $\\\\sum_{i=1}^{100} (\\\\sum_{j=1}^{100} a_{ij} 10^{100-j}) 10^{100-i} = \\\\sum_{i,j} a_{ij} 10^{100-j} 10^{100-i} = \\\\sum_{i,j} a_{ij} 10^{200-i-j}$.\\nThe expressions are identical!\\nSo $\\\\sum_{j=1}^{100} K_j \\\\lambda_j \\\\equiv \\\\sum_{i=1}^{100} R_i \\\\lambda_i \\\\pmod m$.\\nWait, the powers in $K_j$ are based on row index $i$ ($10^{100-i}$), and multiplier is based on col index $j$ ($\\\\lambda_j$).\\nWait, in my notation above: $K_j = \\\\sum_i a_{ij} \\\\lambda_i$. Multiplier is $\\\\lambda_j$?\\nLet's recheck.\\n$K_j$ value is $\\\\sum_{i=1}^{100} a_{ij} 10^{100-i}$. Let's call the weights for $K_j$ as $\\\\mu_i = 10^{100-i}$.\\nSo $K_j = \\\\sum_i a_{ij} \\\\mu_i$.\\n$R_i = \\\\sum_j a_{ij} \\\\nu_j$, where $\\\\nu_j = 10^{100-j}$.\\nNotice $\\\\mu_i = \\\\nu_i$. Let's call it $\\\\omega_k = 10^{100-k}$.\\nSo $K_j = \\\\sum_i a_{ij} \\\\omega_i$.\\n$R_i = \\\\sum_j a_{ij} \\\\omega_j$.\\nThe identity $\\\\sum_{j=1}^{100} K_j \\\\omega_j = \\\\sum_{i=1}^{100} R_i \\\\omega_i$ implies:\\n$\\\\sum_j (\\\\sum_i a_{ij} \\\\omega_i) \\\\omega_j = \\\\sum_{i,j} a_{ij} \\\\omega_i \\\\omega_j$.\\n$\\\\sum_i (\\\\sum_j a_{ij} \\\\omega_j) \\\\omega_i = \\\\sum_{i,j} a_{ij} \\\\omega_i \\\\omega_j$.\\nThey are equal.\\nSo $\\\\sum_{j=1}^{100} K_j \\\\omega_j \\\\equiv \\\\sum_{i=1}^{100} R_i \\\\omega_i \\\\pmod m$.\\nLet $W(m) = \\\\sum_{k=1}^{100} \\\\omega_k m_k$? No.\\nLet $Val(K) = \\\\sum K_j \\\\omega_j$ and $Val(R) = \\\\sum R_i \\\\omega_i$.\\nThey are congruent mod $m$.\\nWe have $R_i, K_j$ divisible by 2013?\\nIf $R_i \\\\equiv 0$ and $K_j \\\\equiv 0$, then $0 \\\\equiv 0$.\\nIf exactly one failure $X$.\\nCase 1: $X = R_k$.\\nThen $R_k \\\\not\\\\equiv 0 \\\\pmod{2013}$. Others are $0$.\\n$R_i \\\\equiv 0 \\\\pmod{2013}$ for $i \\\\neq k$.\\n$K_j \\\\equiv 0 \\\\pmod{2013}$ for all $j$.\\nSo $Val(K) \\\\equiv \\\\sum_j 0 \\\\cdot \\\\omega_j \\\\equiv 0 \\\\pmod{2013}$.\\n$Val(R) \\\\equiv R_k \\\\omega_k + \\\\sum_{i \\\\neq k} R_i \\\\omega_i \\\\equiv R_k \\\\omega_k \\\\pmod{2013}$.\\nSo $R_k \\\\omega_k \\\\equiv 0 \\\\pmod{2013}$.\\nSince $R_k$ is not divisible by 2013, we must have $\\\\omega_k$ invertible modulo factors of 2013.\\nSpecifically, we need $\\\\gcd(\\\\omega_k, 2013) = 1$.\\nLet's check.\\n$\\\\omega_k = 10^{100-k}$.\\nFactors of 2013 are 3, 11, 61.\\nIs $10$ divisible by 3, 11, 61?\\n10 is not div by 3.\\n10 is not div by 11.\\n10 is not div by 61.\\nSo $10^N$ is never 0 mod $p$ for $p \\\\in \\\\{3, 11, 61\\\\}$.\\nThus $\\\\omega_k$ is a unit in $\\\\mathbb{Z}_{2013}$ for all $k$.\\nTherefore $R_k \\\\equiv 0 \\\\pmod{2013}$.\\nThis contradicts the assumption that $R_k$ is not divisible.\\nSo Case 1 is impossible.\\n\\nCase 2: $X = C_k$ (column).\\n$K_k \\\\not\\\\equiv 0 \\\\pmod{2013}$.\\n$R_i \\\\equiv 0$ for all $i$.\\n$K_j \\\\equiv 0$ for $j \\\\neq k$.\\n$Val(K) = K_k \\\\omega_k + \\\\sum_{j \\\\neq k} 0 \\\\equiv K_k \\\\omega_k \\\\pmod{2013}$.\\n$Val(R) = \\\\sum_i 0 \\\\equiv 0$.\\nSo $K_k \\\\omega_k \\\\equiv 0 \\\\pmod{2013}$.\\nSince $\\\\omega_k$ is a unit, $K_k \\\\equiv 0 \\\\pmod{2013}$.\\nContradiction.\\n\\nWait. Is it this simple?\\nLet's double check everything.\\nDefinitions:\\nTable $a_{ij}$.\\nRow $i$: $R_i = \\\\sum_{j=1}^{100} a_{ij} 10^{100-j}$.\\nColumn $j$: $C_j = \\\\sum_{i=1}^{100} a_{ij} 10^{100-i}$.\\nLet $\\\\omega_k = 10^{100-k}$ for $k=1,\\\\dots,100$.\\nThen $R_i = \\\\sum_{j=1}^{100} a_{ij} \\\\omega_j$.\\nAnd $C_j = \\\\sum_{i=1}^{100} a_{ij} \\\\omega_i$.\\nCalculate $\\\\sum_{j=1}^{100} C_j \\\\omega_j$.\\nSubstitute $C_j$:\\n$\\\\sum_{j=1}^{100} (\\\\sum_{i=1}^{100} a_{ij} \\\\omega_i) \\\\omega_j = \\\\sum_{j=1}^{100} \\\\sum_{i=1}^{100} a_{ij} \\\\omega_i \\\\omega_j$.\\nCalculate $\\\\sum_{i=1}^{100} R_i \\\\omega_i$.\\nSubstitute $R_i$:\\n$\\\\sum_{i=1}^{100} (\\\\sum_{j=1}^{100} a_{ij} \\\\omega_j) \\\\omega_i = \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{ij} \\\\omega_j \\\\omega_i$.\\nThey are indeed identical sums.\\nSo $\\\\sum C_j \\\\omega_j = \\\\sum R_i \\\\omega_i$. This holds over integers.\\nNow, modulo $m = 2013$.\\nWe are given that at least 199 of $\\\\{R_i\\\\} \\\\cup \\\\{C_j\\\\}$ are divisible by 2013.\\nLet $F = \\\\{ Y \\\\in \\\\{R_i\\\\} \\\\cup \\\\{C_j\\\\} : Y \\\\not\\\\equiv 0 \\\\pmod{2013} \\\\}$.\\nGiven $|F| \\\\le 1$.\\nSuppose $F = \\\\{R_k\\\\}$. Then $R_k \\\\not\\\\equiv 0$ and $R_i \\\\equiv 0 (i \\\\ne k)$, $C_j \\\\equiv 0$.\\nThen $\\\\sum R_i \\\\omega_i \\\\equiv R_k \\\\omega_k \\\\pmod{2013}$.\\n$\\\\sum C_j \\\\omega_j \\\\equiv 0 \\\\pmod{2013}$.\\nEquality implies $R_k \\\\omega_k \\\\equiv 0 \\\\pmod{2013}$.\\nSince $\\\\omega_k = 10^{100-k}$, $\\\\gcd(\\\\omega_k, 2013) = 1$ because $2013$ has prime factors 3, 11, 61, and 10 is coprime to all of them.\\nSo $R_k \\\\equiv 0 \\\\pmod{2013}$.\\nContradiction.\\nSuppose $F = \\\\{C_k\\\\}$. Then $K_k \\\\not\\\\equiv 0$ and others 0.\\nThen $\\\\sum C_j \\\\omega_j \\\\equiv C_k \\\\omega_k \\\\equiv 0$.\\nContradiction.\\n\\nSo it seems the proof is complete.\\nWhy did I worry about mod 3, 11, 61 separately?\\nJust to double check if I missed anything subtle.\\nThe \\\"weights\\\" $10^{100-k}$ form a geometric progression.\\nIs there any catch?\\nMaybe the definition of \\\"integer formed by digits\\\"?\\nProblem says \\\"100-digit integer\\\". Leading zeros?\\n\\\"digits are nonzero\\\". So $a_{ij} \\\\in \\\\{1, \\\\dots, 9\\\\}$.\\nThis ensures $R_i$ is exactly represented by sum of $a_{ij} 10^{100-j}$.\\nWait, \\\"nonzero digits\\\" means $a_{ij} \\\\ne 0$.\\nWait, the problem statement says:\\n\\\"$10000$ nonzero digits are written in a $100$-by-$100$ table\\\".\\nSo $a_{ij} \\\\in \\\\{1, \\\\dots, 9\\\\}$.\\nThis confirms $a_{ij} \\\\neq 0$.\\nWait, does \\\"nonzero digits\\\" mean just that they are not 0? Or does it imply something about the integer value?\\nUsually, digits are $0..9$. Nonzero means $1..9$.\\nThis ensures that $R_i$ does not start with 0. But that doesn't affect the value formula $\\\\sum a_{ij} 10^{100-j}$.\\nEven with leading zeros, the value is the same. The problem statement \\\"forms a 100-digit integer\\\" might imply no leading zeros allowed, or just that it is treated as a string of 100 digits. With nonzero digits, it's definitely a 100-digit number without leading zero issues.\\n\\nThe reasoning seems solid.\\nLet me just re-read carefully to ensure no misinterpretation.\\n\\\"prove that if at least 199 of these 200 numbers are divisible by 2013, then all of them are divisible by 2013.\\\"\\nMy proof shows that if exactly 199 are divisible, then the non-divisible one must be divisible too.\\nSo if 200 are divisible, obviously all are.\\nSo the condition \\\"at least 199\\\" covers the cases $199, 200$.\\nIf 199 are div, 1 is not. We showed this is impossible.\\nThus 200 must be div.\\nWait, if 0 are divisible? No, \\\"at least 199\\\".\\nSo the proof holds.\\n\\nLet's check if there are any edge cases or flaws.\\n1. Are the bases $10$ and exponents $100-i$ correct?\\nRow $i$: digits $a_{i1}, a_{i2}, \\\\dots, a_{i100}$.\\nValue $\\\\sum_{j=1}^{100} a_{ij} 10^{100-j}$.\\nExponent for $a_{i1}$ is $10^{99}$. For $a_{i100}$ is $10^0$.\\nFormula $\\\\sum_{j=1}^{100} a_{ij} \\\\omega_j$ with $\\\\omega_j = 10^{100-j}$.\\nCorrect.\\nColumn $j$: digits $a_{1j}, a_{2j}, \\\\dots, a_{100j}$.\\nValue $\\\\sum_{i=1}^{100} a_{ij} 10^{100-i}$.\\nExponent for $a_{1j}$ is $10^{99}$. For $a_{100,j}$ is $10^0$.\\nFormula $\\\\sum_{i=1}^{100} a_{ij} \\\\omega_i$.\\nCorrect.\\nIdentity $\\\\sum_j C_j \\\\omega_j = \\\\sum_{i,j} a_{ij} \\\\omega_i \\\\omega_j$.\\n$\\\\sum_i R_i \\\\omega_i = \\\\sum_{i,j} a_{ij} \\\\omega_j \\\\omega_i$.\\nBoth are $\\\\sum_{i,j} a_{ij} \\\\omega_i \\\\omega_j$.\\nThis is perfectly symmetric and correct.\\nCongruence modulo $m$.\\n$\\\\sum C_j \\\\omega_j \\\\equiv \\\\sum R_i \\\\omega_i \\\\pmod m$.\\nIf we have 199 multiples of $m$.\\nLet $S_R = \\\\sum R_i \\\\omega_i \\\\pmod m$.\\nLet $S_C = \\\\sum C_j \\\\omega_j \\\\pmod m$.\\n$S_R = S_C$.\\nIf $R_k$ is the only non-multiple. Then $R_k \\\\not\\\\equiv 0$.\\nOther $R_i \\\\equiv 0$, all $C_j \\\\equiv 0$.\\n$S_R \\\\equiv R_k \\\\omega_k \\\\pmod m$.\\n$S_C \\\\equiv 0 \\\\pmod m$.\\n$R_k \\\\omega_k \\\\equiv 0 \\\\pmod m$.\\n$\\\\omega_k = 10^{100-k}$.\\n$m = 2013 = 3 \\\\times 11 \\\\times 61$.\\n$\\\\gcd(10, 3) = 1 \\\\implies \\\\gcd(10^{100-k}, 3) = 1$.\\n$\\\\gcd(10, 11) = 1 \\\\implies \\\\gcd(10^{100-k}, 11) = 1$.\\n$\\\\gcd(10, 61) = 1 \\\\implies \\\\gcd(10^{100-k}, 61) = 1$.\\nSo $\\\\omega_k$ is invertible modulo 2013.\\nThus $R_k \\\\equiv 0 \\\\pmod{2013}$.\\nContradiction.\\nSame for $C_k$.\\n\\nIs there any case where this argument fails?\\nOnly if $\\\\omega_k$ is not coprime to 2013.\\nThis happens if $10$ shares a factor with 2013.\\n$2013 = 3 \\\\times 11 \\\\times 61$.\\n$10 = 2 \\\\times 5$.\\nCommon factors: None.\\nSo $\\\\omega_k$ is always invertible.\\n\\nIs it possible that \\\"nonzero digits\\\" allows digits to be larger than 9? \\\"digit\\\" usually implies base 10 symbol. \\\"nonzero digits\\\" usually means $1..9$. If digits were from a higher base, it wouldn't be called \\\"digits\\\" in a standard sense without qualification, and \\\"divisible by 2013\\\" would refer to the integer value. The context of 100-digit integer strongly implies base 10. Even if digits were symbols, the algebraic expansion $a_{ij} 10^p$ assumes interpretation in base 10.\\nWait, what if $a_{ij}$ represents the digit at position $j$ in base $B$? The problem says \\\"100-digit integer\\\". This means base 10.\\nSo digits are coefficients of $10^k$.\\nIf the digits were just integers $a_{ij}$, the construction \\\"forms an integer\\\" implies $0, 1, \\\\dots, 9$.\\nWith \\\"nonzero\\\", $1, \\\\dots, 9$.\\nSo the assumption holds.\\n\\nOne small detail: \\\"not necessarily distinct\\\". Irrelevant.\\nIs there any subtlety about $N(100)$ vs $N(1)$? The order is left-to-right, top-to-bottom.\\nIndices $i, j$ are fixed.\\nSeems robust.\\n\\nWait, I should check the problem source or similar problems to ensure I'm not missing a \\\"catch\\\".\\nThis looks like a variation of \\\"Magic Squares\\\" or \\\"Latin Square\\\" divisibility problems, often solved using the determinant or sum-product identities.\\nThe identity used here is essentially $\\\\det(M) = 0$ for matrix of all ones? No.\\nIt's just linear dependence of row/column polynomials evaluated at roots?\\nLet $P_i(x) = \\\\sum a_{ij} x^{100-j}$. Then $R_i = P_i(10)$.\\nLet $Q_j(y) = \\\\sum a_{ij} y^{100-i}$. Then $C_j = Q_j(10)$? No.\\nLet's stick to the indices.\\nThe identity $\\\\sum R_i x^i = \\\\sum C_j y^j$ is not quite right.\\nIt relies on the specific symmetry of powers $10^{100-j}$ and $10^{100-i}$.\\nIf the base was different, say $a_{ij}$ contributed $B^j$ in row and $B^i$ in column, the cross terms would match.\\nHere row contribution to $a_{ij}$ is $10^{100-j}$, column is $10^{100-i}$.\\nWeighted sum $\\\\sum R_i W_i$ matches $\\\\sum C_j W_j$ iff $W_i 10^{100-j} \\\\sim W_j 10^{100-i}$.\\nIf we chose $W_i = 10^{100-i}$, then $W_i \\\\cdot (\\\\text{coeff of } R_i) = 10^{100-i} \\\\cdot 10^{100-j}$.\\nFor column side, $W_j \\\\cdot (\\\\text{coeff of } C_j) = 10^{100-j} \\\\cdot 10^{100-i}$.\\nMatches perfectly.\\nSo the choice of weights $10^{100-k}$ is dictated by the structure.\\nThis structure is robust.\\n\\nAre there any counterexamples?\\nSuppose $m=3$.\\n$2013$ is not $3$.\\nSuppose $m=11$.\\nWeights $10^{100-k} \\\\equiv (-1)^{100-k} \\\\equiv (-1)^k \\\\pmod{11}$.\\nInvertibility: $10$ is not divisible by 11. So invertible.\\nIf we had $m=2$. $10 \\\\equiv 0 \\\\pmod 2$.\\nThen $\\\\omega_k \\\\equiv 0$.\\nThen $\\\\sum C_j \\\\omega_j \\\\equiv 0$ and $\\\\sum R_i \\\\omega_i \\\\equiv 0$ becomes $0=0$.\\nIf $m=2$, and digits are 0 or 1.\\nRows/cols sum of digits? No.\\nIntegers are values. If $m=2$, $R_i$ parity.\\nParity of $\\\\sum a_{ij} 10^{100-j}$.\\nIf $100-j \\\\ge 1$, term is even.\\nOnly $j=100$ term $a_{i,100} 10^0 = a_{i,100}$ affects parity.\\nSo $R_i \\\\equiv a_{i,100} \\\\pmod 2$.\\n$C_j \\\\equiv a_{100,j} \\\\pmod 2$.\\nSum $\\\\sum C_j 0 \\\\equiv 0$, $\\\\sum R_i 0 \\\\equiv 0$.\\nIf $m$ divides 10.\\nOur $m=2013$ does not divide 10.\\nSo $\\\\gcd(m, 10)=1$ is crucial.\\n$2013$ ends in 3, so coprime to 2 and 5.\\nSo $\\\\gcd(10, 2013) = 1$.\\nSo $\\\\omega_k$ is indeed a unit.\\n\\nThe proof seems complete and correct.\\nJust write it down clearly.\\n\\nSteps:\\n1. Define $R_i$ and $C_j$ using formulas.\\n2. Note that $R_i = \\\\sum_{j=1}^{100} a_{ij} \\\\omega_j$ and $C_j = \\\\sum_{i=1}^{100} a_{ij} \\\\omega_i$ with $\\\\omega_k = 10^{100-k}$.\\n3. Construct the linear combination identity: $\\\\sum_{j=1}^{100} C_j \\\\omega_j = \\\\sum_{i=1}^{100} R_i \\\\omega_i$.\\n Proof: Expand both sides into $\\\\sum_{i,j} a_{ij} \\\\omega_i \\\\omega_j$.\\n4. Work modulo $m = 2013$.\\n5. Given that at least 199 numbers are divisible by $m$.\\n Let $F$ be the set of numbers not divisible by $m$. $|F| \\\\le 1$.\\n If $F = \\\\emptyset$, trivial.\\n Assume $|F|=1$. Let $X \\\\in F$.\\n Show $X$ cannot exist.\\n Split into two subcases: $X = R_k$ or $X = C_k$.\\n6. Case $X = R_k$:\\n $R_i \\\\equiv 0$ for $i \\\\ne k$, $C_j \\\\equiv 0$ for all $j$.\\n Substitute into identity: $\\\\sum C_j \\\\omega_j \\\\equiv 0$, $\\\\sum R_i \\\\omega_i \\\\equiv R_k \\\\omega_k$.\\n Get $R_k \\\\omega_k \\\\equiv 0 \\\\pmod m$.\\n Argue $\\\\gcd(\\\\omega_k, m) = 1$ since $m=3 \\\\cdot 11 \\\\cdot 61$ and $\\\\gcd(10, m)=1$.\\n Conclude $R_k \\\\equiv 0 \\\\pmod m$, contradiction.\\n7. Case $X = C_k$: Similar derivation leads to $C_k \\\\equiv 0 \\\\pmod m$, contradiction.\\n8. Conclude $F = \\\\emptyset$. Thus all 200 numbers are divisible.\\n\\nDouble check the \\\"nonzero digits\\\" part again.\\nDoes it play any role?\\nIt ensures $a_{ij} \\\\in \\\\{1, \\\\dots, 9\\\\}$.\\nIt prevents $a_{ij}=0$.\\nDoes $a_{ij}=0$ change anything?\\nIf $a_{ij}$ could be 0, the algebraic sum $\\\\sum a_{ij} \\\\omega_i \\\\omega_j$ still holds.\\nThe value of the integer formed by digits is still the sum.\\nSo the proof works even if zeros were allowed.\\nMaybe the problem adds \\\"nonzero\\\" to simplify visualization or ensure the numbers are \\\"proper\\\" 100-digit integers (no leading zeros).\\nActually, if $a_{i1}=0$, the row forms an integer with fewer than 100 digits (leading zero). The problem states \\\"forms a 100-digit integer\\\". This implies $a_{i1} \\\\ne 0$.\\nSince all are nonzero, this is satisfied.\\nSo the condition is naturally handled.\\n\\nIs there any dependency on \\\"at least 199\\\"?\\nIf the threshold was lower, say 99, we couldn't conclude.\\nIf threshold was 200, it's tautology.\\nWhy 199? Because it guarantees that we have enough information to pin down the \\\"weight\\\" factor for the remaining 1.\\nSpecifically, if we had 2 failures, say $R_1$ and $C_1$, we would have $R_1 \\\\omega_1 + R_2 \\\\omega_2 + \\\\dots \\\\equiv C_1 \\\\omega_1 + \\\\dots$\\nThen $R_1 \\\\omega_1 - C_1 \\\\omega_1 + \\\\dots \\\\equiv 0$.\\n$(R_1 - C_1) \\\\omega_1 \\\\equiv \\\\sum_{j \\\\ne 1} C_j \\\\omega_j - \\\\sum_{i \\\\ne 1} R_i \\\\omega_i$.\\nIf $C_j, R_i$ are 0 for $j,i \\\\ne 1$, then $(R_1 - C_1) \\\\omega_1 \\\\equiv 0$.\\nSo $R_1 \\\\equiv C_1 \\\\pmod m$.\\nIf we have 2 failures, they must be congruent mod $m$.\\nBut they are not necessarily non-zero.\\nWait, if $R_1, C_1$ are not 0, and all others 0.\\nThen $R_1 \\\\omega_1 \\\\equiv C_1 \\\\omega_1 \\\\implies R_1 \\\\equiv C_1$.\\nBut we don't know if they are non-zero. They could be non-zero and equal.\\nIf $R_1 = C_1 = 2013/2$ (not integer).\\nSay $R_1 = 2014$. Divisible? No.\\nIs it possible to construct a table with 198 divisible by 2013, and $R_1, C_1$ such that $R_1 \\\\equiv C_1 \\\\not\\\\equiv 0$?\\nIf so, then \\\"at least 199\\\" is the tight bound.\\nIf we had 2 failures, it would be consistent if $R_1 \\\\equiv C_1 \\\\pmod{2013}$ (and no other terms).\\nActually, the identity is $\\\\sum C_j \\\\omega_j - \\\\sum R_i \\\\omega_i = 0$.\\nLet's test the hypothesis: Can we have 2 failures $R_1, C_1$ with $R_1 \\\\not\\\\equiv 0, C_1 \\\\not\\\\equiv 0, R_1 \\\\equiv C_1 \\\\pmod m$?\\nAnd $R_i \\\\equiv 0, C_j \\\\equiv 0$ otherwise.\\nWe need to construct digits $a_{ij}$ satisfying the linear system for remainders.\\nThis is under-determined (10000 vars, 2 constraints plus integrality).\\nSo yes, likely possible to have 2 failures.\\nSo the statement \\\"if at least 199 ... then all\\\" is sharp.\\nBut proving it is impossible for 199 is sufficient.\\nMy proof did that.\\n\\nWait, check the \\\"at least 199\\\" phrasing.\\nIf 199 are divisible, then the remaining 1 is not (since if it were, we'd have 200).\\nMy proof assumes \\\"exactly 1 is not\\\".\\nIf \\\"at least 199\\\" includes the case \\\"200\\\", that's trivial.\\nSo the critical case is \\\"exactly 199 divisible\\\". Which means \\\"exactly 1 not divisible\\\".\\nSo the logic holds.\\n\\nFinal check on arithmetic.\\n$2013 = 3 \\\\times 11 \\\\times 61$.\\n$10 \\\\equiv 1 \\\\pmod 3$.\\n$10 \\\\equiv -1 \\\\pmod{11}$.\\n$10 \\\\equiv 10 \\\\pmod{61}$.\\nUnits group orders.\\n$\\\\phi(2013) = 2 \\\\times 10 \\\\times 60 = 1200$.\\n$10^{1200} \\\\equiv 1 \\\\pmod{2013}$.\\nThis doesn't matter, just invertibility.\\n$\\\\gcd(10, 3)=1, \\\\gcd(10, 11)=1, \\\\gcd(10, 61)=1$.\\nConfirmed.\\n\\nStructure of the write-up:\\n1. Formulate row and column integers.\\n2. Establish the identity $\\\\sum_{i=1}^{100} R_i \\\\omega_i = \\\\sum_{j=1}^{100} C_j \\\\omega_j$.\\n3. Apply modulo 2013.\\n4. Use the \\\"at most 1 failure\\\" condition.\\n5. Deduce that the unique failure must be divisible by 2013 due to invertibility of weights.\\n6. Conclude.\\n\\nOne detail: $R_i = \\\\sum_{j=1}^{100} a_{ij} 10^{100-j}$.\\nIndices $j=1..100$.\\n$a_{i1}$ is most significant digit. Power $10^{99}$.\\nLet's check indices in sum.\\n$\\\\sum_{j=1}^{100} a_{ij} \\\\omega_j$.\\n$\\\\omega_j = 10^{100-j}$.\\n$j=1 \\\\implies 10^{99}$. Correct.\\n$j=100 \\\\implies 10^0$. Correct.\\nSame for columns. $C_j = \\\\sum_{i=1}^{100} a_{ij} 10^{100-i}$.\\n$i=1 \\\\implies 10^{99}$. Correct.\\n$i=100 \\\\implies 10^0$. Correct.\\n\\nThe solution seems perfect.\\n\\nIs there any ambiguity about \\\"from left to right\\\" vs index?\\nUsually matrices are indexed $i$ (rows) top to bottom, $j$ (cols) left to right.\\nRow $i$ reads $a_{i1} \\\\dots a_{i100}$.\\nCol $j$ reads $a_{1j} \\\\dots a_{100j}$.\\nThis matches my formulation.\\n\\nI will write down the proof following these steps.\\nNo special \\\"tricks\\\" needed beyond the basic linear combination.\\nJustification for every step will include the explicit summations.\\n\\nFinal check:\\nProblem asks to prove.\\n\\\"If at least 199... then all...\\\"\\nProof:\\nLet $S$ be the set of 200 numbers.\\nAssume subset $K \\\\subset S$ has size $\\\\ge 199$ such that $\\\\forall x \\\\in K, 2013 | x$.\\nIf $|K|=200$, done.\\nAssume $|K|=199$. Then there exists $x \\\\in S \\\\setminus K$ such that $2013 \\\\nmid x$.\\nDerive contradiction.\\nDone.\\n\\nIs it possible that $2013$ is not $3 \\\\times 671$?\\n$3 \\\\times 671 = 2013$.\\n$671 = 11 \\\\times 61$.\\nCorrect.\\nIs 671 prime? No, $660 = 60 \\\\times 11$.\\n$671 / 11 = 61$. Correct.\\n\\nOkay, ready to write.\\n\\nWait, could I mention the generalization?\\n\\\"For any integer $m$ coprime to 10...\\\".\\nThis makes the solution more elegant but might distract from the specific question. I'll stick to the specific numbers but noting coprimality is good style.\\nActually, mentioning 2013 specifically is safer.\\n\\nOne thought:\\nCould $\\\\omega_i$ be defined as $10^{i}$?\\nIf $R_i = \\\\sum a_{ij} 10^j$? No, that changes the value.\\nUsing $\\\\omega_i = 10^{100-i}$ is the natural choice matching the \\\"place value\\\" of digits.\\nIf I used reverse indexing $x_k = 10^k$, then $R_i = \\\\sum a_{ij} x_{100-j}$.\\nIt gets messy. Current definition is cleanest.\\n\\nThe result seems to rely heavily on $\\\\sum R_i 10^{100-i} = \\\\sum C_j 10^{100-j}$? No.\\nThe identity is $\\\\sum R_i 10^{100-i} = \\\\sum C_j 10^{100-j}$.\\nLet's check this again.\\nLHS: $\\\\sum_i (\\\\sum_j a_{ij} 10^{100-j}) 10^{100-i} = \\\\sum_{i,j} a_{ij} 10^{200-i-j}$.\\nRHS: $\\\\sum_j (\\\\sum_i a_{ij} 10^{100-i}) 10^{100-j} = \\\\sum_{i,j} a_{ij} 10^{100-i} 10^{100-j} = \\\\sum_{i,j} a_{ij} 10^{200-i-j}$.\\nYes.\\nThis identity holds for ANY field/ring.\\nSo for modulo 2013.\\nThe invertibility of $10^{100-i}$ modulo 2013 is the key.\\nIf 2013 shared a factor with 10, say 2.\\nThen $10^{200-i-j}$ terms are divisible by 2.\\nWait. If $m=2$, $2013$ replaced by 2.\\nThen $\\\\sum R_i 2^{\\\\dots} \\\\equiv \\\\sum C_j 2^{\\\\dots}$.\\nTerms are $a_{ij} 2^{\\\\dots}$.\\nThe lowest power is $10^0 \\\\cdot 10^0 = 1$.\\nWait. $10^k$ is not 0 mod 2 for $k \\\\ge 1$.\\nBut $10 \\\\equiv 0 \\\\pmod 2$.\\nSo $10^{100-i}$ is 0 mod 2 for $i \\\\le 99$.\\nFor $i=100$, it is $1 \\\\pmod 2$.\\nSo the \\\"invertibility\\\" fails only for specific indices if $m$ shares factors with base.\\nBut here $m=2013$ and base 10 are coprime.\\nSo invertibility holds for all $i$.\\n\\nWait, $10^{100-i}$ is invertible mod 2013.\\n$\\\\gcd(10^{100-i}, 2013) = 1$.\\nYes.\\nSo no index-dependent issues.\\nEverything is uniform.\\n\\nOne check on \\\"divisible by 2013\\\".\\nDoes it imply the number is $0 \\\\pmod{2013}$? Yes.\\nIf a number is negative? Digits are positive.\\nSo number is positive.\\nDivisibility is straightforward.\\n\\nOkay, plan is solid.\\n\\nOne minor point:\\nCan I simply say \\\"Let $x = 10^{100-i}$\\\"? No, exponents vary.\\nI will use $w_i = 10^{100-i}$.\\nIdentity: $\\\\sum_{i=1}^{100} R_i w_i = \\\\sum_{j=1}^{100} C_j w_j$.\\nNote indices: Row $i$ multiplied by weight depending on row index $i$ (matches exponent).\\nColumn $j$ multiplied by weight depending on col index $j$? No, wait.\\nIn LHS: $\\\\sum_i R_i w_i$. $w_i = 10^{100-i}$.\\nIn RHS: $\\\\sum_j C_j w_j$. $w_j = 10^{100-j}$.\\nWait, in my derivation:\\nLHS of expansion: $\\\\sum_{i,j} a_{ij} 10^{100-i} 10^{100-j}$.\\nRHS: $\\\\sum_{i,j} a_{ij} 10^{100-j} 10^{100-i}$.\\nWait, the expression $\\\\sum C_j w_j$ expands to $\\\\sum_j (\\\\sum_i a_{ij} 10^{100-i}) 10^{100-j}$.\\nThis involves $10^{100-i}$ from $C_j$ and $10^{100-j}$ from $w_j$.\\nProduct is $10^{200-i-j}$.\\nExpression $\\\\sum R_i w_i$ expands to $\\\\sum_i (\\\\sum_j a_{ij} 10^{100-j}) 10^{100-i}$.\\nThis involves $10^{100-j}$ from $R_i$ and $10^{100-i}$ from $w_i$.\\nProduct is $10^{200-i-j}$.\\nThe terms match.\\nNote that in the first expansion, the multiplier $w_j$ is associated with column $j$.\\nIn the second expansion, multiplier $w_i$ is associated with row $i$.\\nSo the identity is $\\\\sum_{i=1}^{100} R_i \\\\cdot 10^{100-i} = \\\\sum_{j=1}^{100} C_j \\\\cdot 10^{100-j}$.\\nWait, the RHS indices $j$ run $1..100$.\\nLet's call the weights $\\\\tau_k = 10^{100-k}$.\\nThen Identity is $\\\\sum_{i=1}^{100} R_i \\\\tau_i = \\\\sum_{j=1}^{100} C_j \\\\tau_j$.\\nWait, check the index set for $\\\\tau$.\\nFor row $i$, we multiply by $\\\\tau_i = 10^{100-i}$.\\nFor column $j$, we multiply by $\\\\tau_j = 10^{100-j}$.\\nThe set of values $\\\\{\\\\tau_1, \\\\dots, \\\\tau_{100}\\\\}$ is the same.\\nSo the sum is over the set of indices.\\nThis is slightly confusing to read as $\\\\sum R_i \\\\tau_i$ vs $\\\\sum C_j \\\\tau_j$.\\nUsually one writes $\\\\sum R_i w_i$.\\nBut in the RHS sum, $j$ is the dummy variable for columns.\\nSo it is $\\\\sum_{j=1}^{100} C_j \\\\tau_j$.\\nSo the weights are applied with the same functional form $\\\\tau_k = 10^{100-k}$ to the appropriate index.\\nWait.\\nLet's check if the term $a_{ij}$ in $R_i$ gets multiplied by $\\\\tau_i$.\\n$R_i = \\\\sum_j a_{ij} 10^{100-j}$.\\nTerm is $a_{ij} 10^{100-j}$.\\nMultiply by $\\\\tau_i = 10^{100-i}$.\\nResult $a_{ij} 10^{100-j} 10^{100-i}$.\\nIn $C_j = \\\\sum_i a_{ij} 10^{100-i}$.\\nTerm is $a_{ij} 10^{100-i}$.\\nMultiply by $\\\\tau_j = 10^{100-j}$.\\nResult $a_{ij} 10^{100-i} 10^{100-j}$.\\nThey match.\\nSo the identity is indeed:\\n$$ \\\\sum_{i=1}^{100} R_i 10^{100-i} = \\\\sum_{j=1}^{100} C_j 10^{100-j} $$\\nIs this correct?\\nLet's verify the \\\"indices\\\" of the multiplication.\\nLHS: $\\\\sum R_i \\\\times (\\\\text{factor based on } i)$.\\nRHS: $\\\\sum C_j \\\\times (\\\\text{factor based on } j)$.\\nBut the factor for row $i$ is the same function of $i$ as the factor for col $j$.\\nFunction $f(k) = 10^{100-k}$.\\nSo $\\\\sum R_i f(i) = \\\\sum C_j f(j)$.\\nYes.\\n\\nWait, looking back at my scratchpad.\\nI wrote $\\\\sum_{j} C_j \\\\omega_j = \\\\sum_{i} R_i \\\\omega_i$.\\nHere $\\\\omega_k = 10^{100-k}$.\\nIn the identity I verified just now:\\nLHS term for $R_i$ is $R_i \\\\cdot 10^{100-i} = R_i \\\\omega_i$.\\nRHS term for $C_j$ is $C_j \\\\cdot 10^{100-j} = C_j \\\\omega_j$.\\nSo the identity is $\\\\sum_i R_i \\\\omega_i = \\\\sum_j C_j \\\\omega_j$.\\nThis matches.\\n\\nWait, in my previous thought process I wrote:\\n$LHS: \\\\sum C_j \\\\omega_j$.\\n$RHS: \\\\sum R_i \\\\omega_i$.\\nLet's trace carefully which is which.\\n$\\\\sum_{j=1}^{100} C_j \\\\omega_j = \\\\sum_{j=1}^{100} (\\\\sum_{i=1}^{100} a_{ij} \\\\omega_i) \\\\omega_j = \\\\sum_{i,j} a_{ij} \\\\omega_i \\\\omega_j$.\\n$\\\\sum_{i=1}^{100} R_i \\\\omega_i = \\\\sum_{i=1}^{100} (\\\\sum_{j=1}^{100} a_{ij} \\\\omega_j) \\\\omega_i = \\\\sum_{i,j} a_{ij} \\\\omega_j \\\\omega_i$.\\nSince $\\\\omega_i \\\\omega_j$ is commutative, sums are equal.\\nSo $\\\\sum_{i=1}^{100} R_i \\\\omega_i = \\\\sum_{j=1}^{100} C_j \\\\omega_j$.\\nYes, this is the correct identity.\\n\\nOne confusion could arise from the indices.\\n$R_i$ is row $i$. $\\\\omega_i$ uses $i$.\\n$C_j$ is col $j$. $\\\\omega_j$ uses $j$.\\nThis is consistent.\\nWait, $R_i = \\\\sum a_{ij} \\\\omega_j$. (Power depends on column index $j$).\\n$C_j = \\\\sum a_{ij} \\\\omega_i$. (Power depends on row index $i$).\\nSo the identity $\\\\sum R_i \\\\omega_i$ weights row $i$ by $\\\\omega_i = 10^{100-i}$.\\nThis aligns with the power in $C_j$.\\nWait, $R_i$ has $\\\\omega_j$ inside. Multiplying by $\\\\omega_i$ creates $\\\\omega_i \\\\omega_j$.\\n$C_j$ has $\\\\omega_i$ inside. Multiplying by $\\\\omega_j$ creates $\\\\omega_j \\\\omega_i$.\\nSo yes, it balances perfectly.\\nThe logic holds.\\n\\nJust a sanity check on the problem wording:\\n\\\"From left to right, each row forms a 100-digit integer.\\\"\\nLeftmost digit is $a_{i1}$. Rightmost is $a_{i100}$.\\nValue $d_{1} d_{2} \\\\dots d_{100} = \\\\sum d_k 10^{100-k}$.\\nHere $d_1 = a_{i1}$. So $\\\\sum a_{ij} 10^{100-j}$.\\nThis confirms $100-j$ exponent.\\nAlso \\\"top to bottom\\\". Top digit $a_{1j}$.\\nValue $\\\\sum a_{ij} 10^{100-i}$.\\nThis confirms $100-i$ exponent.\\nAnd $\\\\omega_k = 10^{100-k}$.\\nIdentity confirmed.\\n\\nThe solution is ready.\\n\\n\\n# Output\\nRespond with exactly one lowercase word and nothing else -- no punctuation or\\nexplanation:\\n- For an incomplete prefix, use `yes` with probability equal to the expected\\n normalized rubric credit of a fresh continuation.\\n- For a complete response, use `yes` with probability equal to its realized\\n normalized rubric credit.\\n- Otherwise use `no`.\\n\"\n },\n {\n \"role\": \"assistant\",\n \"content\": \"yes\"\n }\n]","isConversational":true},"enable_thinking":{"kind":"bool","value":false,"string":"false"},"id":{"kind":"string","value":"8cf1cb3370041166"},"source":{"kind":"string","value":"fineproofs"},"sample_index":{"kind":"number","value":1,"string":"1"},"step_idx":{"kind":"number","value":45,"string":"45"},"num_steps":{"kind":"number","value":45,"string":"45"},"label":{"kind":"string","value":"yes"},"correct":{"kind":"bool","value":true,"string":"true"},"loss_weight":{"kind":"number","value":1,"string":"1"},"reward":{"kind":"number","value":0.9955357142857143,"string":"0.995536"}}},{"rowIdx":98,"cells":{"messages":{"kind":"list like","value":[{"role":"user","content":"# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefix is incomplete, imagine starting exactly where it ends and sampling\none fresh continuation from the same fixed policy and sampling procedure used to\ncollect the rollout data. The completed response will be graded like an olympiad\nproof: partial credit is awarded only for correct, rigorous, and clearly justified\nprogress, and the awarded points are normalized by the maximum available points.\nTreat your probability of `yes` as the expected normalized rubric credit in\nthe interval from 0 to 1.\n\nIf the prefix is a complete response, judge its realized outcome as written. Treat\nyour probability of `yes` as that response's realized normalized rubric credit in\nthe interval from 0 to 1. Do not imagine replacing or continuing an already\ncomplete response.\n\nThis is an eventual rubric-value forecast, not merely a check that the reasoning\nshown so far looks plausible. A valid prefix can still have low value when the\nremaining proof is difficult. A flawed prefix can have nonzero value when the\npolicy can recover. A guessed final answer without a valid proof earns little\ncredit unless the grading scheme says otherwise.\n\nReference attempts, when present, are side information for estimating the policy\nand problem. They are not part of the fixed prefix and are not visible to the\nhypothetical continuation.\n\n# Problem\n$10000$ nonzero digits are written in a $100$ -by- $100$ table, one digit per cell. From left to right, each row forms a $100$ -digit integer. From top to bottom, each column forms a $100$ -digit integer. So the rows and columns form $200$ integers (each with $100$ digits), not necessarily distinct. Prove that if at least $199$ of these $200$ numbers are divisible by $2013$ , then all of them are divisible by $2013$ .\n\n# Reference attempts\n## Attempt 1\nLet $n = 100$ and $M = 2013$.\nLet $d_{i,j}$ be the digit in the $i$-th row and $j$-th column, where $1 \\le i, j \\le n$. The problem states that $d_{i,j} \\in \\{1, 2, \\dots, 9\\}$.\nThe integer formed by the $i$-th row is denoted by $R_i$. Based on the convention of writing digits from left to right, $R_i$ is given by:\n$$ R_i = \\sum_{j=1}^{n} d_{i,j} 10^{n-j} $$\nSimilarly, the integer formed by the $j$-th column is denoted by $C_j$:\n$$ C_j = \\sum_{i=1}^{n} d_{i,j} 10^{n-i} $$\nWe are given that at least $2n-1$ of these $2n$ integers ($R_1, \\dots, R_n, C_1, \\dots, C_n$) are divisible by $M$. We must prove that all of them are divisible by $M$.\n\nConsider the following algebraic identity involving the row and column sums:\n$$ S = \\sum_{i=1}^{n} R_i 10^{n-i} $$\nSubstituting the expression for $R_i$:\n$$ S = \\sum_{i=1}^{n} 10^{n-i} \\left( \\sum_{j=1}^{n} d_{i,j} 10^{n-j} \\right) = \\sum_{i=1}^{n} \\sum_{j=1}^{n} d_{i,j} 10^{2n - i - j} $$\nSimilarly, consider the sum of the column integers weighted by powers of 10:\n$$ T = \\sum_{j=1}^{n} C_j 10^{n-j} $$\nSubstituting the expression for $C_j$:\n$$ T = \\sum_{j=1}^{n} 10^{n-j} \\left( \\sum_{i=1}^{n} d_{i,j} 10^{n-i} \\right) = \\sum_{j=1}^{n} \\sum_{i=1}^{n} d_{i,j} 10^{2n - i - j} $$\nSince the summation variables $i$ and $j$ run over the same range $[1, n]$, the double sums are identical:\n$$ \\sum_{i=1}^{n} \\sum_{j=1}^{n} d_{i,j} 10^{2n - i - j} = \\sum_{j=1}^{n} \\sum_{i=1}^{n} d_{i,j} 10^{2n - i - j} $$\nThus, we have the identity:\n$$ \\sum_{i=1}^{n} R_i 10^{n-i} \\equiv \\sum_{j=1}^{n} C_j 10^{n-j} \\pmod M $$\nThis identity holds as an equality of integers, and consequently as a congruence modulo $M$.\n\nLet $S_R = \\{ i \\in \\{1, \\dots, n\\} \\mid R_i \\equiv 0 \\pmod M \\}$ and $S_C = \\{ j \\in \\{1, \\dots, n\\} \\mid C_j \\equiv 0 \\pmod M \\}$.\nThe condition that at least $199$ of the $200$ numbers are divisible by $M$ implies that $|S_R| + |S_C| \\ge 199$.\nSince $|S_R| \\le 100$ and $|S_C| \\le 100$, the possible values for $(|S_R|, |S_C|)$ are limited to cases where the number of non-divisible integers is at most 1 (since $200 - 199 = 1$). Specifically, the number of indices not in $S_R \\cup S_C$ is $k$ such that $200 - |S_R| - |S_C| \\ge 200 - 199 = 1$, so the count of non-divisible numbers is $N_{bad} = 200 - |S_R| - |S_C| \\le 1$.\n\nWe examine the possible scenarios for the set of \"bad\" (non-divisible) numbers:\n1. **Zero bad numbers:** $|S_R| = 100$ and $|S_C| = 100$. In this case, all 200 numbers are divisible by $M$, and we are done.\n2. **One bad number:** There are two subcases due to the symmetry between rows and columns.\n * *Subcase 2a:* $R_i \\equiv 0 \\pmod M$ for all $i$, but $C_n \\not\\equiv 0 \\pmod M$ (so $|S_R|=100, |S_C|=99$).\n Applying the identity modulo $M$:\n $$ \\sum_{i=1}^{n} R_i 10^{n-i} \\equiv \\sum_{j=1}^{n} C_j 10^{n-j} \\pmod M $$\n The Left Hand Side (LHS) is:\n $$ \\sum_{i=1}^{n} 0 \\cdot 10^{n-i} \\equiv 0 \\pmod M $$\n The Right Hand Side (RHS) is:\n $$ \\sum_{j=1}^{n-1} C_j 10^{n-j} + C_n 10^{n-n} $$\n Since $C_1, \\dots, C_{n-1}$ are divisible by $M$, the first part is $0$. The term $10^{n-n} = 10^0 = 1$.\n So RHS $\\equiv C_n \\pmod M$.\n Thus, $0 \\equiv C_n \\pmod M$, which implies $C_n$ is divisible by $M$. This contradicts the assumption that $C_n$ is bad. Therefore, this case is impossible.\n * *Subcase 2b:* $C_j \\equiv 0 \\pmod M$ for all $j$, but $R_n \\not\\equiv 0 \\pmod M$ (so $|S_R|=99, |S_C|=100$).\n By symmetry with Subcase 2a, we have:\n $$ \\sum_{i=1}^{n} R_i 10^{n-i} \\equiv \\sum_{j=1}^{n} C_j 10^{n-j} \\pmod M $$\n $$ R_n \\cdot 10^0 + 0 \\equiv 0 \\implies R_n \\equiv 0 \\pmod M $$\n Again, this is a contradiction.\n\nAny other configuration would involve at least 2 bad numbers (e.g., $R_n$ and $C_n$ both bad, or $R_a$ and $R_b$ both bad), resulting in at most $200-2 = 198$ divisible numbers, which violates the hypothesis.\n\nConsequently, the only possible configuration consistent with the hypothesis is the first one: all 200 numbers are divisible by $M$. Thus, the claim is proven.\n\n**Answer:**\nThe integers formed by the rows are $R_i = \\sum_{j=1}^{100} d_{i,j} 10^{100-j}$ and the integers formed by the columns are $C_j = \\sum_{i=1}^{100} d_{i,j} 10^{100-i}$. Consider the sums:\n$$ \\sum_{i=1}^{100} R_i 10^{100-i} \\quad \\text{and} \\quad \\sum_{j=1}^{100} C_j 10^{100-j}. $$\nExpanding these sums reveals that both are equal to $\\sum_{i=1}^{100} \\sum_{j=1}^{100} d_{i,j} 10^{200-i-j}$. Hence,\n$$ \\sum_{i=1}^{100} R_i 10^{100-i} \\equiv \\sum_{j=1}^{100} C_j 10^{100-j} \\pmod{2013}. $$\nAssume that at least 199 of the numbers are divisible by 2013. This implies that at most one number is not divisible. If there are no non-divisible numbers, we are done. If there is exactly one non-divisible number, symmetry allows us to assume it is a column integer $C_n$ (the other case is symmetric to considering $R_n$). Then $R_i \\equiv 0 \\pmod{2013}$ for all $i$ and $C_j \\equiv 0 \\pmod{2013}$ for all $j < n$. Substituting these into the identity yields:\n$$ \\sum_{i=1}^{100} 0 \\cdot 10^{100-i} \\equiv \\sum_{j=1}^{n-1} 0 \\cdot 10^{100-j} + C_n \\cdot 10^0 \\pmod{2013} $$\n$$ 0 \\equiv C_n \\pmod{2013}. $$\nThis implies $C_n$ is divisible by 2013, contradicting the assumption that it was non-divisible. Therefore, it is impossible to have exactly one non-divisible number. The hypothesis implies that all 200 numbers must be divisible by 2013.\n\n## Attempt 2\nLet the digits in the table be denoted by $a_{ij}$ for $1 \\le i, j \\le 100$, where $a_{ij} \\in \\{1, 2, \\dots, 9\\}$ are the nonzero digits.\nLet $R_i$ be the integer formed by the $i$-th row, and $C_j$ be the integer formed by the $j$-th column.\nBased on the positional notation in base 10, we can express these integers as follows:\n$$ R_i = \\sum_{j=1}^{100} a_{ij} 10^{100-j} $$\n$$ C_j = \\sum_{i=1}^{100} a_{ij} 10^{100-i} $$\nLet us define a sequence of weights $w_k = 10^{100-k}$ for $k = 1, 2, \\dots, 100$.\nThen the expressions become:\n$$ R_i = \\sum_{j=1}^{100} a_{ij} w_j $$\n$$ C_j = \\sum_{i=1}^{100} a_{ij} w_i $$\nWe consider the quantity $S = \\sum_{i=1}^{100} R_i w_i$. Substituting the expression for $R_i$, we get:\n$$ \\begin{aligned} S &= \\sum_{i=1}^{100} w_i \\left( \\sum_{j=1}^{100} a_{ij} w_j \\right) \\\\ &= \\sum_{i=1}^{100} \\sum_{j=1}^{100} a_{ij} w_i w_j \\end{aligned} $$\nSimilarly, consider the quantity $T = \\sum_{j=1}^{100} C_j w_j$. Substituting the expression for $C_j$, we get:\n$$ \\begin{aligned} T &= \\sum_{j=1}^{100} w_j \\left( \\sum_{i=1}^{100} a_{ij} w_i \\right) \\\\ &= \\sum_{j=1}^{100} \\sum_{i=1}^{100} a_{ij} w_j w_i \\end{aligned} $$\nSince finite sums commute and multiplication of integers is commutative ($w_i w_j = w_j w_i$), we observe that $S = T$. That is:\n$$ \\sum_{i=1}^{100} R_i w_i = \\sum_{j=1}^{100} C_j w_j $$\nThis equality holds over the integers. Therefore, it also holds modulo any integer $m$. In particular, let $m = 2013$.\n$$ \\sum_{i=1}^{100} R_i w_i \\equiv \\sum_{j=1}^{100} C_j w_j \\pmod{2013} $$\nWe are given that at least 199 of the 200 integers $\\{R_1, \\dots, R_{100}, C_1, \\dots, C_{100}\\}$ are divisible by 2013. This means there is at most 1 integer among them that is not divisible by 2013.\nWe proceed by contradiction. Suppose there exists exactly one integer $X$ in the set which is not divisible by 2013. There are two cases:\nCase 1: The non-divisible integer is a row, say $R_k$ for some $k \\in \\{1, \\dots, 100\\}$.\nThen $R_i \\equiv 0 \\pmod{2013}$ for all $i \\neq k$, and $C_j \\equiv 0 \\pmod{2013}$ for all $j \\in \\{1, \\dots, 100\\}$.\nSubstituting these into our congruence relation:\n$$ \\sum_{i=1}^{100} R_i w_i \\equiv R_k w_k + \\sum_{i \\neq k} 0 \\cdot w_i \\equiv R_k w_k \\pmod{2013} $$\n$$ \\sum_{j=1}^{100} C_j w_j \\equiv \\sum_{j=1}^{100} 0 \\cdot w_j \\equiv 0 \\pmod{2013} $$\nEquating these results gives:\n$$ R_k w_k \\equiv 0 \\pmod{2013} $$\nRecall that $w_k = 10^{100-k}$. The modulus is $2013$. We factorize 2013 as $2013 = 3 \\times 11 \\times 61$.\nSince $10$ is not divisible by 3, 11, or 61 (obvious inspection or calculation), $\\gcd(10, 2013) = 1$. Consequently, $\\gcd(w_k, 2013) = \\gcd(10^{100-k}, 2013) = 1$.\nThis implies that $w_k$ is invertible modulo 2013. Multiplying both sides of $R_k w_k \\equiv 0 \\pmod{2013}$ by the modular inverse of $w_k$, we obtain:\n$$ R_k \\equiv 0 \\pmod{2013} $$\nThis contradicts the assumption that $R_k$ is not divisible by 2013.\n\nCase 2: The non-divisible integer is a column, say $C_k$ for some $k \\in \\{1, \\dots, 100\\}$.\nThen $R_i \\equiv 0 \\pmod{2013}$ for all $i$, and $C_j \\equiv 0 \\pmod{2013}$ for all $j \\neq k$.\nSubstituting these into the congruence relation:\n$$ \\sum_{i=1}^{100} R_i w_i \\equiv 0 \\pmod{2013} $$\n$$ \\sum_{j=1}^{100} C_j w_j \\equiv C_k w_k + \\sum_{j \\neq k} 0 \\cdot w_j \\equiv C_k w_k \\pmod{2013} $$\nEquating these gives:\n$$ C_k w_k \\equiv 0 \\pmod{2013} $$\nAs before, since $\\gcd(w_k, 2013) = 1$, it follows that $C_k \\equiv 0 \\pmod{2013}$.\nThis contradicts the assumption that $C_k$ is not divisible by 2013.\n\nSince both cases lead to a contradiction, our initial assumption that there is exactly one non-divisible integer must be false.\nThus, there are either 0 or fewer than 1 non-divisible integers. Given that \"at least 199\" are divisible, it is impossible to have fewer than 199 divisible. If exactly 1 failed, we have a contradiction. Thus, 0 fail.\nTherefore, all 200 integers $R_1, \\dots, R_{100}$ and $C_1, \\dots, C_{100}$ are divisible by 2013.\n\n## Attempt 3\nThe problem asks us to prove that if in a $100 \\times 100$ table of nonzero digits, at least 199 of the 200 integers formed by rows and columns are divisible by 2013, then all 200 integers are divisible by 2013.\n\nLet the digits in the table be $a_{i,j}$ for $1 \\le i, j \\le 100$, where $a_{i,j} \\in \\{1, \\dots, 9\\}$.\nLet $R_i$ be the integer represented by the $i$-th row, and $C_j$ be the integer represented by the $j$-th column.\n$$R_i = \\sum_{j=1}^{100} a_{i,j} 10^{100-j}, \\quad C_j = \\sum_{i=1}^{100} a_{i,j} 10^{100-i}.$$\nThe set of 200 integers is $S = \\{R_1, \\dots, R_{100}, C_1, \\dots, C_{100}\\}$.\nWe are given that at least 199 elements of $S$ are divisible by 2013. This implies that at most 1 element of $S$ is not divisible by 2013.\n\nFirst, we factorize 2013:\n$$2013 = 3 \\times 11 \\times 61.$$\nSince 3, 11, and 61 are pairwise coprime, an integer is divisible by 2013 if and only if it is divisible by 3, 11, and 61. Thus, it suffices to show that under the given condition, every integer in $S$ is divisible by 3, by 11, and by 61.\nLet $p$ be a prime factor of 2013. We will show that if at most 1 element of $S$ is not divisible by $p$, then all elements of $S$ are divisible by $p$.\nThe prime factors are 3, 11, and 61. Notice that none of these primes divide 10, so $10$ is invertible modulo $p$.\nLet $x_k = 10^{100-k} \\pmod p$ for $k=1, \\dots, 100$. Since $p \\nmid 10$, $x_k \\not\\equiv 0 \\pmod p$ for all $k$.\nWe can write $R_i$ and $C_j$ as:\n$$R_i \\equiv \\sum_{j=1}^{100} a_{i,j} x_j \\pmod p,$$\n$$C_j \\equiv \\sum_{i=1}^{100} a_{i,j} x_i \\pmod p.$$\n(Note: We use the same sequence $x$ for both row and column coefficients because $10^{100-j}$ and $10^{100-i}$ are just indexed terms of the sequence $10^k$).\n\nConsider the sum $W = \\sum_{i=1}^{100} x_i R_i \\pmod p$. Substituting the expressions for $R_i$:\n$$W \\equiv \\sum_{i=1}^{100} x_i \\left( \\sum_{j=1}^{100} a_{i,j} x_j \\right) \\pmod p.$$\nBy changing the order of summation:\n$$W \\equiv \\sum_{j=1}^{100} x_j \\left( \\sum_{i=1}^{100} a_{i,j} x_i \\right) \\pmod p.$$\nThe inner sum $\\sum_{i=1}^{100} a_{i,j} x_i$ is exactly $C_j$. Thus:\n$$W \\equiv \\sum_{j=1}^{100} x_j C_j \\pmod p.$$\nSo we have the identity:\n$$\\sum_{i=1}^{100} x_i R_i \\equiv \\sum_{j=1}^{100} x_j C_j \\pmod p.$$\n\nNow we use the hypothesis that at most 1 number in $S$ is not divisible by $p$.\nSuppose there is exactly one number in $S$ that is not divisible by $p$.\nThere are two cases:\nCase 1: The non-divisible number is a row $R_k$.\nThen all $R_i$ for $i \\ne k$ are divisible by $p$, and all $C_j$ for all $j$ are divisible by $p$.\nSubstituting into the identity:\n$$\\sum_{i=1}^{100} x_i R_i \\equiv x_k R_k + \\sum_{i \\ne k} x_i \\cdot 0 \\equiv x_k R_k \\pmod p.$$\n$$\\sum_{j=1}^{100} x_j C_j \\equiv \\sum_{j=1}^{100} x_j \\cdot 0 \\equiv 0 \\pmod p.$$\nFrom the identity, $x_k R_k \\equiv 0 \\pmod p$.\nSince $x_k = 10^{100-k}$ is not divisible by $p$ (as $p \\nmid 10$), we must have $R_k \\equiv 0 \\pmod p$. This contradicts the assumption that $R_k$ is not divisible by $p$.\n\nCase 2: The non-divisible number is a column $C_l$.\nThen all $R_i$ for all $i$ are divisible by $p$, and all $C_j$ for $j \\ne l$ are divisible by $p$.\nSubstituting into the identity:\n$$\\sum_{i=1}^{100} x_i R_i \\equiv \\sum_{i=1}^{100} x_i \\cdot 0 \\equiv 0 \\pmod p.$$\n$$\\sum_{j=1}^{100} x_j C_j \\equiv x_l C_l + \\sum_{j \\ne l} x_j \\cdot 0 \\equiv x_l C_l \\pmod p.$$\nFrom the identity, $x_l C_l \\equiv 0 \\pmod p$.\nAgain, since $x_l \\not\\equiv 0 \\pmod p$, we must have $C_l \\equiv 0 \\pmod p$. This contradicts the assumption that $C_l$ is not divisible by $p$.\n\nThus, in both cases, the assumption that a number is not divisible by $p$ leads to a contradiction. Therefore, all numbers in $S$ must be divisible by $p$.\nThis holds for each prime factor $p \\in \\{3, 11, 61\\}$. Since the numbers are divisible by each of 3, 11, and 61, they are divisible by $3 \\times 11 \\times 61 = 2013$.\nTherefore, all 200 integers are divisible by 2013.\n\n**Final Answer:**\nThe problem asks to prove that all 200 numbers are divisible by 2013.\nLet $S = \\{R_1, \\dots, R_{100}, C_1, \\dots, C_{100}\\}$ be the set of the 200 integers. We are given that at least 199 elements of $S$ are divisible by 2013. This means at most 1 element is not divisible by 2013.\nSince $2013 = 3 \\times 11 \\times 61$, we consider each prime factor $p \\in \\{3, 11, 61\\}$ separately. For any such $p$, if $x \\equiv 0 \\pmod{2013}$, then $x \\equiv 0 \\pmod p$. Thus, at most 1 element of $S$ is not divisible by $p$.\nLet $x_k = 10^{100-k} \\pmod p$. Since $p \\nmid 10$, $x_k \\not\\equiv 0 \\pmod p$ for all $k$.\nWe observe the following congruence relation modulo $p$:\n$$ \\sum_{i=1}^{100} x_i R_i = \\sum_{i=1}^{100} x_i \\sum_{j=1}^{100} a_{i,j} x_j = \\sum_{j=1}^{100} x_j \\sum_{i=1}^{100} a_{i,j} x_i = \\sum_{j=1}^{100} x_j C_j. $$\nIf there were an element not divisible by $p$, say $R_k$, then all $C_j$ and $R_i$ ($i \\ne k$) are divisible by $p$. The relation implies $x_k R_k \\equiv \\sum x_j C_j \\equiv 0 \\pmod p$, so $R_k \\equiv 0 \\pmod p$, a contradiction. Similarly, if $C_l$ were the only non-divisible element, all $R_i$ and $C_j$ ($j \\ne l$) are divisible by $p$. The relation implies $x_l C_l \\equiv \\sum x_i R_i \\equiv 0 \\pmod p$, so $C_l \\equiv 0 \\pmod p$, a contradiction.\nTherefore, no element can be not divisible by $p$. Thus, all 200 elements are divisible by 3, 11, and 61. Being divisible by these coprime factors, all 200 elements are divisible by $3 \\times 11 \\times 61 = 2013$.\n\nThe final result is that all of them are divisible by 2013.\n\n## Attempt 4\nLet $N = 2013$. We are given a $100 \\times 100$ table of nonzero digits $d_{i,j} \\in \\{1, \\dots, 9\\}$ for $1 \\le i, j \\le 100$.\nLet $R_i$ denote the integer formed by the $i$-th row and $C_j$ denote the integer formed by the $j$-th column. Specifically, the digits are concatenated from left-to-right for rows and top-to-bottom for columns. Thus:\n$$R_i = \\sum_{j=1}^{100} d_{i,j} 10^{100-j}$$\n$$C_j = \\sum_{i=1}^{100} d_{i,j} 10^{100-i}$$\nWe are given that at least 199 of these 200 integers are divisible by $N$. We aim to prove that all 200 integers are divisible by $N$.\n\nFirst, observe that $\\gcd(10, 2013) = 1$. Thus, any power of 10, denoted $\\omega_k = 10^{100-k}$, is coprime to $2013$ and has a multiplicative inverse modulo $2013$.\nThe problem states that for at least 199 indices pairs $(i,j)$ associated with the 200 integers, the value is $0 \\pmod{2013}$. Let $\\mathcal{I}$ be the set of indices of the integers that are divisible by $N$. We have $|\\mathcal{I}| \\ge 199$.\nThe set of all integers consists of $\\{R_1, \\dots, R_{100}\\} \\cup \\{C_1, \\dots, C_{100}\\}$.\nThe number of non-divisible integers is $200 - |\\mathcal{I}| \\le 1$.\nLet $n_r$ be the number of rows not divisible by $N$ and $n_c$ be the number of columns not divisible by $N$. The total number of non-divisible integers is $n_r + n_c \\le 1$.\nThis leaves two possibilities to consider (excluding the trivial case where 0 are non-divisible, in which case the theorem holds):\n1. Exactly one row fails to be divisible ($n_r = 1$) and all columns are divisible ($n_c = 0$).\n2. Exactly one column fails to be divisible ($n_c = 1$) and all rows are divisible ($n_r = 0$).\n\nBy symmetry (the transpose of the table swaps rows and columns but preserves the structure and divisibility properties), we only need to disprove case 1. If case 1 leads to a contradiction, case 2 leads to the same contradiction by relabeling rows and columns.\n\n**Disproving Case 1:**\nAssume that there exists an index $k \\in \\{1, \\dots, 100\\}$ such that $R_k$ is not divisible by $N$, while for all $i \\neq k$, $R_i$ is divisible by $N$. Furthermore, assume that all columns $C_j$ are divisible by $N$.\n\nThe condition that all columns are divisible means that for every $j \\in \\{1, \\dots, 100\\}$:\n$$C_j = \\sum_{i=1}^{100} d_{i,j} 10^{100-i} \\equiv 0 \\pmod{2013}$$\nLet us denote the weight associated with the $m$-th position in a number as $\\omega_m = 10^{100-m}$. Then the conditions are:\n1. For all $j$, $\\sum_{i=1}^{100} d_{i,j} \\omega_i \\equiv 0 \\pmod{2013}$.\n2. For all $i \\neq k$, $\\sum_{j=1}^{100} d_{i,j} \\omega_j \\equiv 0 \\pmod{2013}$.\n\nConsider the equation for column $j$ (from condition 1):\n$$d_{k,j} \\omega_k + \\sum_{i \\neq k} d_{i,j} \\omega_i \\equiv 0 \\pmod{2013}$$\nSince $\\omega_k$ is a unit modulo $2013$ (as $\\gcd(10, 2013)=1$), we can isolate $d_{k,j}$:\n$$d_{k,j} \\equiv -\\omega_k^{-1} \\sum_{i \\neq k} d_{i,j} \\omega_i \\pmod{2013}$$\n\nNow, consider the expression for $R_k$ (modulo $N$). By condition 2 (applied to $R_k$ itself, though we only know it's non-divisible, we can still manipulate the expression to check its value):\n$$R_k = \\sum_{j=1}^{100} d_{k,j} \\omega_j$$\nSubstitute the expression for $d_{k,j}$ into this sum:\n$$R_k \\equiv \\sum_{j=1}^{100} \\left( -\\omega_k^{-1} \\sum_{i \\neq k} d_{i,j} \\omega_i \\right) \\omega_j \\pmod{2013}$$\nRearranging the summation order:\n$$R_k \\equiv -\\omega_k^{-1} \\sum_{i \\neq k} \\omega_i \\left( \\sum_{j=1}^{100} d_{i,j} \\omega_j \\right) \\pmod{2013}$$\nThe inner sum $\\sum_{j=1}^{100} d_{i,j} \\omega_j$ is precisely the integer value of row $i$, denoted $R_i$.\nThus, we have:\n$$R_k \\equiv -\\omega_k^{-1} \\sum_{i \\neq k} \\omega_i R_i \\pmod{2013}$$\nUnder our assumption for Case 1, $R_i$ is divisible by $N$ for all $i \\neq k$. Therefore, $R_i \\equiv 0 \\pmod{2013}$.\nSubstituting $R_i \\equiv 0$ into the equation:\n$$R_k \\equiv -\\omega_k^{-1} \\sum_{i \\neq k} \\omega_i \\cdot 0 \\equiv 0 \\pmod{2013}$$\nThis implies that $R_k$ is divisible by $2013$. This contradicts the assumption that $R_k$ is not divisible by $2013$.\n\nTherefore, it is impossible for exactly one row to fail while all columns succeed. By symmetry, it is also impossible for exactly one column to fail while all rows succeed.\n\nConsequently, the only way to satisfy the condition that at least 199 of the 200 integers are divisible by 2013 is if none of them fail. That is, $n_r = 0$ and $n_c = 0$.\n\nThus, all 200 integers are divisible by 2013.\n\n**Proof that no other answer works (rigorous summary):**\nWe established that $n_r + n_c \\le 1$ is required by the problem statement.\n1. If $n_r + n_c = 0$, the condition is met trivially.\n2. If $n_r + n_c = 1$, we proved it leads to a contradiction ($R_k \\equiv 0 \\pmod{2013}$ if assumed non-zero, or $C_m \\equiv 0 \\pmod{2013}$ if assumed non-zero).\nSince the case $n_r + n_c > 1$ yields fewer than 199 divisible numbers (specifically, at most 198), it is excluded by the problem hypothesis.\nThus, the only valid state is where all 200 integers are divisible.\n\nQ.E.D.\n\nThe final result is that all 200 numbers must be divisible by 2013.\n\nFinal Answer: All 200 of the integers are divisible by 2013.\n\n# Fixed response prefix\nLet the $100 \\times 100$ table be denoted by $A = (a_{i,j})$ where $1 \\le i, j \\le 100$.\nThe digits are nonzero, so $a_{i,j} \\in \\{1, 2, \\dots, 9\\}$. Wait, the problem says \"nonzero digits\".\nThis means $a_{i,j} \\in \\{1, \\dots, 9\\}$? Or could it be that they are just integers $1, \\dots, 9$? \"Digits\" usually implies base 10 digits. Nonzero means $\\{1, \\dots, 9\\}$.\n\nLet $R_i$ be the integer formed by the $i$-th row for $1 \\le i \\le 100$.\nLet $C_j$ be the integer formed by the $j$-th column for $1 \\le j \\le 100$.\nWe have a set of $N = 200$ integers $\\{R_1, \\dots, R_{100}, C_1, \\dots, C_{100}\\}$.\nLet $S$ be this set (as a multiset of values). The size is 200.\nWe are given that at least 199 of these numbers are divisible by $2013$.\nWe need to prove that all of them are divisible by $2013$.\nThis is equivalent to: If $| \\{ x \\in S : x \\equiv 0 \\pmod{2013} \\} | \\ge 199$, then for all $x \\in S$, $x \\equiv 0 \\pmod{2013}$.\n\nLet's check the modulus. $2013 = 3 \\times 11 \\times 61$.\nWait, $2+0+1+3 = 6$, so divisible by 3. $2013 / 3 = 671$.\n$671 = 11 \\times 61$. So $2013 = 3 \\times 11 \\times 61$.\nThis factorization might not be crucial immediately, but good to know.\n\nLet's denote $r(i)$ as the remainder of $R_i$ modulo 2013, and $c(j)$ as the remainder of $C_j$ modulo 2013.\nThere are $m=100$ rows and $n=100$ columns.\nLet $r_i \\equiv R_i \\pmod m$ and $c_j \\equiv C_j \\pmod m$. Actually the modulus is $M=2013$.\nSo $R_i = \\sum_{j=1}^{100} a_{i,j} 10^{100-j}$ and $C_j = \\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\nThe condition is that $r_i = 0$ or $c_j = 0$ for almost all indices.\nSuppose exactly one number is not divisible by 2013. Without loss of generality, assume $R_k$ is not divisible, and all other $R_i$ ($i \\ne k$) and all $C_j$ ($1 \\le j \\le 100$) are divisible.\nOr maybe two are not divisible, etc. But we want to show that the assumption of $\\ge 199$ implies all are divisible. So let's try to construct a contradiction if there is one non-divisible number.\nOr rather, let's analyze the sum of remainders or relations between $R_i$'s and $C_j$'s.\nSpecifically, $R_i$ depends on the row entries, $C_j$ depends on column entries. They intersect at $(i,j)$.\n\n# Output\nRespond with exactly one lowercase word and nothing else -- no punctuation or\nexplanation:\n- For an incomplete prefix, use `yes` with probability equal to the expected\n normalized rubric credit of a fresh continuation.\n- For a complete response, use `yes` with probability equal to its realized\n normalized rubric credit.\n- Otherwise use `no`.\n"},{"role":"assistant","content":"yes"}],"string":"[\n {\n \"role\": \"user\",\n \"content\": \"# Role\\nYou are forecasting the rubric value of a response trajectory for an olympiad\\nmathematics proof. The problem and response prefix below are fixed. The prefix may\\nbe incomplete, or it may already contain the complete response including its final\\nanswer.\\n\\nIf the prefix is incomplete, imagine starting exactly where it ends and sampling\\none fresh continuation from the same fixed policy and sampling procedure used to\\ncollect the rollout data. The completed response will be graded like an olympiad\\nproof: partial credit is awarded only for correct, rigorous, and clearly justified\\nprogress, and the awarded points are normalized by the maximum available points.\\nTreat your probability of `yes` as the expected normalized rubric credit in\\nthe interval from 0 to 1.\\n\\nIf the prefix is a complete response, judge its realized outcome as written. Treat\\nyour probability of `yes` as that response's realized normalized rubric credit in\\nthe interval from 0 to 1. Do not imagine replacing or continuing an already\\ncomplete response.\\n\\nThis is an eventual rubric-value forecast, not merely a check that the reasoning\\nshown so far looks plausible. A valid prefix can still have low value when the\\nremaining proof is difficult. A flawed prefix can have nonzero value when the\\npolicy can recover. A guessed final answer without a valid proof earns little\\ncredit unless the grading scheme says otherwise.\\n\\nReference attempts, when present, are side information for estimating the policy\\nand problem. They are not part of the fixed prefix and are not visible to the\\nhypothetical continuation.\\n\\n# Problem\\n$10000$ nonzero digits are written in a $100$ -by- $100$ table, one digit per cell. From left to right, each row forms a $100$ -digit integer. From top to bottom, each column forms a $100$ -digit integer. So the rows and columns form $200$ integers (each with $100$ digits), not necessarily distinct. Prove that if at least $199$ of these $200$ numbers are divisible by $2013$ , then all of them are divisible by $2013$ .\\n\\n# Reference attempts\\n## Attempt 1\\nLet $n = 100$ and $M = 2013$.\\nLet $d_{i,j}$ be the digit in the $i$-th row and $j$-th column, where $1 \\\\le i, j \\\\le n$. The problem states that $d_{i,j} \\\\in \\\\{1, 2, \\\\dots, 9\\\\}$.\\nThe integer formed by the $i$-th row is denoted by $R_i$. Based on the convention of writing digits from left to right, $R_i$ is given by:\\n$$ R_i = \\\\sum_{j=1}^{n} d_{i,j} 10^{n-j} $$\\nSimilarly, the integer formed by the $j$-th column is denoted by $C_j$:\\n$$ C_j = \\\\sum_{i=1}^{n} d_{i,j} 10^{n-i} $$\\nWe are given that at least $2n-1$ of these $2n$ integers ($R_1, \\\\dots, R_n, C_1, \\\\dots, C_n$) are divisible by $M$. We must prove that all of them are divisible by $M$.\\n\\nConsider the following algebraic identity involving the row and column sums:\\n$$ S = \\\\sum_{i=1}^{n} R_i 10^{n-i} $$\\nSubstituting the expression for $R_i$:\\n$$ S = \\\\sum_{i=1}^{n} 10^{n-i} \\\\left( \\\\sum_{j=1}^{n} d_{i,j} 10^{n-j} \\\\right) = \\\\sum_{i=1}^{n} \\\\sum_{j=1}^{n} d_{i,j} 10^{2n - i - j} $$\\nSimilarly, consider the sum of the column integers weighted by powers of 10:\\n$$ T = \\\\sum_{j=1}^{n} C_j 10^{n-j} $$\\nSubstituting the expression for $C_j$:\\n$$ T = \\\\sum_{j=1}^{n} 10^{n-j} \\\\left( \\\\sum_{i=1}^{n} d_{i,j} 10^{n-i} \\\\right) = \\\\sum_{j=1}^{n} \\\\sum_{i=1}^{n} d_{i,j} 10^{2n - i - j} $$\\nSince the summation variables $i$ and $j$ run over the same range $[1, n]$, the double sums are identical:\\n$$ \\\\sum_{i=1}^{n} \\\\sum_{j=1}^{n} d_{i,j} 10^{2n - i - j} = \\\\sum_{j=1}^{n} \\\\sum_{i=1}^{n} d_{i,j} 10^{2n - i - j} $$\\nThus, we have the identity:\\n$$ \\\\sum_{i=1}^{n} R_i 10^{n-i} \\\\equiv \\\\sum_{j=1}^{n} C_j 10^{n-j} \\\\pmod M $$\\nThis identity holds as an equality of integers, and consequently as a congruence modulo $M$.\\n\\nLet $S_R = \\\\{ i \\\\in \\\\{1, \\\\dots, n\\\\} \\\\mid R_i \\\\equiv 0 \\\\pmod M \\\\}$ and $S_C = \\\\{ j \\\\in \\\\{1, \\\\dots, n\\\\} \\\\mid C_j \\\\equiv 0 \\\\pmod M \\\\}$.\\nThe condition that at least $199$ of the $200$ numbers are divisible by $M$ implies that $|S_R| + |S_C| \\\\ge 199$.\\nSince $|S_R| \\\\le 100$ and $|S_C| \\\\le 100$, the possible values for $(|S_R|, |S_C|)$ are limited to cases where the number of non-divisible integers is at most 1 (since $200 - 199 = 1$). Specifically, the number of indices not in $S_R \\\\cup S_C$ is $k$ such that $200 - |S_R| - |S_C| \\\\ge 200 - 199 = 1$, so the count of non-divisible numbers is $N_{bad} = 200 - |S_R| - |S_C| \\\\le 1$.\\n\\nWe examine the possible scenarios for the set of \\\"bad\\\" (non-divisible) numbers:\\n1. **Zero bad numbers:** $|S_R| = 100$ and $|S_C| = 100$. In this case, all 200 numbers are divisible by $M$, and we are done.\\n2. **One bad number:** There are two subcases due to the symmetry between rows and columns.\\n * *Subcase 2a:* $R_i \\\\equiv 0 \\\\pmod M$ for all $i$, but $C_n \\\\not\\\\equiv 0 \\\\pmod M$ (so $|S_R|=100, |S_C|=99$).\\n Applying the identity modulo $M$:\\n $$ \\\\sum_{i=1}^{n} R_i 10^{n-i} \\\\equiv \\\\sum_{j=1}^{n} C_j 10^{n-j} \\\\pmod M $$\\n The Left Hand Side (LHS) is:\\n $$ \\\\sum_{i=1}^{n} 0 \\\\cdot 10^{n-i} \\\\equiv 0 \\\\pmod M $$\\n The Right Hand Side (RHS) is:\\n $$ \\\\sum_{j=1}^{n-1} C_j 10^{n-j} + C_n 10^{n-n} $$\\n Since $C_1, \\\\dots, C_{n-1}$ are divisible by $M$, the first part is $0$. The term $10^{n-n} = 10^0 = 1$.\\n So RHS $\\\\equiv C_n \\\\pmod M$.\\n Thus, $0 \\\\equiv C_n \\\\pmod M$, which implies $C_n$ is divisible by $M$. This contradicts the assumption that $C_n$ is bad. Therefore, this case is impossible.\\n * *Subcase 2b:* $C_j \\\\equiv 0 \\\\pmod M$ for all $j$, but $R_n \\\\not\\\\equiv 0 \\\\pmod M$ (so $|S_R|=99, |S_C|=100$).\\n By symmetry with Subcase 2a, we have:\\n $$ \\\\sum_{i=1}^{n} R_i 10^{n-i} \\\\equiv \\\\sum_{j=1}^{n} C_j 10^{n-j} \\\\pmod M $$\\n $$ R_n \\\\cdot 10^0 + 0 \\\\equiv 0 \\\\implies R_n \\\\equiv 0 \\\\pmod M $$\\n Again, this is a contradiction.\\n\\nAny other configuration would involve at least 2 bad numbers (e.g., $R_n$ and $C_n$ both bad, or $R_a$ and $R_b$ both bad), resulting in at most $200-2 = 198$ divisible numbers, which violates the hypothesis.\\n\\nConsequently, the only possible configuration consistent with the hypothesis is the first one: all 200 numbers are divisible by $M$. Thus, the claim is proven.\\n\\n**Answer:**\\nThe integers formed by the rows are $R_i = \\\\sum_{j=1}^{100} d_{i,j} 10^{100-j}$ and the integers formed by the columns are $C_j = \\\\sum_{i=1}^{100} d_{i,j} 10^{100-i}$. Consider the sums:\\n$$ \\\\sum_{i=1}^{100} R_i 10^{100-i} \\\\quad \\\\text{and} \\\\quad \\\\sum_{j=1}^{100} C_j 10^{100-j}. $$\\nExpanding these sums reveals that both are equal to $\\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} d_{i,j} 10^{200-i-j}$. Hence,\\n$$ \\\\sum_{i=1}^{100} R_i 10^{100-i} \\\\equiv \\\\sum_{j=1}^{100} C_j 10^{100-j} \\\\pmod{2013}. $$\\nAssume that at least 199 of the numbers are divisible by 2013. This implies that at most one number is not divisible. If there are no non-divisible numbers, we are done. If there is exactly one non-divisible number, symmetry allows us to assume it is a column integer $C_n$ (the other case is symmetric to considering $R_n$). Then $R_i \\\\equiv 0 \\\\pmod{2013}$ for all $i$ and $C_j \\\\equiv 0 \\\\pmod{2013}$ for all $j < n$. Substituting these into the identity yields:\\n$$ \\\\sum_{i=1}^{100} 0 \\\\cdot 10^{100-i} \\\\equiv \\\\sum_{j=1}^{n-1} 0 \\\\cdot 10^{100-j} + C_n \\\\cdot 10^0 \\\\pmod{2013} $$\\n$$ 0 \\\\equiv C_n \\\\pmod{2013}. $$\\nThis implies $C_n$ is divisible by 2013, contradicting the assumption that it was non-divisible. Therefore, it is impossible to have exactly one non-divisible number. The hypothesis implies that all 200 numbers must be divisible by 2013.\\n\\n## Attempt 2\\nLet the digits in the table be denoted by $a_{ij}$ for $1 \\\\le i, j \\\\le 100$, where $a_{ij} \\\\in \\\\{1, 2, \\\\dots, 9\\\\}$ are the nonzero digits.\\nLet $R_i$ be the integer formed by the $i$-th row, and $C_j$ be the integer formed by the $j$-th column.\\nBased on the positional notation in base 10, we can express these integers as follows:\\n$$ R_i = \\\\sum_{j=1}^{100} a_{ij} 10^{100-j} $$\\n$$ C_j = \\\\sum_{i=1}^{100} a_{ij} 10^{100-i} $$\\nLet us define a sequence of weights $w_k = 10^{100-k}$ for $k = 1, 2, \\\\dots, 100$.\\nThen the expressions become:\\n$$ R_i = \\\\sum_{j=1}^{100} a_{ij} w_j $$\\n$$ C_j = \\\\sum_{i=1}^{100} a_{ij} w_i $$\\nWe consider the quantity $S = \\\\sum_{i=1}^{100} R_i w_i$. Substituting the expression for $R_i$, we get:\\n$$ \\\\begin{aligned} S &= \\\\sum_{i=1}^{100} w_i \\\\left( \\\\sum_{j=1}^{100} a_{ij} w_j \\\\right) \\\\\\\\ &= \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{ij} w_i w_j \\\\end{aligned} $$\\nSimilarly, consider the quantity $T = \\\\sum_{j=1}^{100} C_j w_j$. Substituting the expression for $C_j$, we get:\\n$$ \\\\begin{aligned} T &= \\\\sum_{j=1}^{100} w_j \\\\left( \\\\sum_{i=1}^{100} a_{ij} w_i \\\\right) \\\\\\\\ &= \\\\sum_{j=1}^{100} \\\\sum_{i=1}^{100} a_{ij} w_j w_i \\\\end{aligned} $$\\nSince finite sums commute and multiplication of integers is commutative ($w_i w_j = w_j w_i$), we observe that $S = T$. That is:\\n$$ \\\\sum_{i=1}^{100} R_i w_i = \\\\sum_{j=1}^{100} C_j w_j $$\\nThis equality holds over the integers. Therefore, it also holds modulo any integer $m$. In particular, let $m = 2013$.\\n$$ \\\\sum_{i=1}^{100} R_i w_i \\\\equiv \\\\sum_{j=1}^{100} C_j w_j \\\\pmod{2013} $$\\nWe are given that at least 199 of the 200 integers $\\\\{R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}\\\\}$ are divisible by 2013. This means there is at most 1 integer among them that is not divisible by 2013.\\nWe proceed by contradiction. Suppose there exists exactly one integer $X$ in the set which is not divisible by 2013. There are two cases:\\nCase 1: The non-divisible integer is a row, say $R_k$ for some $k \\\\in \\\\{1, \\\\dots, 100\\\\}$.\\nThen $R_i \\\\equiv 0 \\\\pmod{2013}$ for all $i \\\\neq k$, and $C_j \\\\equiv 0 \\\\pmod{2013}$ for all $j \\\\in \\\\{1, \\\\dots, 100\\\\}$.\\nSubstituting these into our congruence relation:\\n$$ \\\\sum_{i=1}^{100} R_i w_i \\\\equiv R_k w_k + \\\\sum_{i \\\\neq k} 0 \\\\cdot w_i \\\\equiv R_k w_k \\\\pmod{2013} $$\\n$$ \\\\sum_{j=1}^{100} C_j w_j \\\\equiv \\\\sum_{j=1}^{100} 0 \\\\cdot w_j \\\\equiv 0 \\\\pmod{2013} $$\\nEquating these results gives:\\n$$ R_k w_k \\\\equiv 0 \\\\pmod{2013} $$\\nRecall that $w_k = 10^{100-k}$. The modulus is $2013$. We factorize 2013 as $2013 = 3 \\\\times 11 \\\\times 61$.\\nSince $10$ is not divisible by 3, 11, or 61 (obvious inspection or calculation), $\\\\gcd(10, 2013) = 1$. Consequently, $\\\\gcd(w_k, 2013) = \\\\gcd(10^{100-k}, 2013) = 1$.\\nThis implies that $w_k$ is invertible modulo 2013. Multiplying both sides of $R_k w_k \\\\equiv 0 \\\\pmod{2013}$ by the modular inverse of $w_k$, we obtain:\\n$$ R_k \\\\equiv 0 \\\\pmod{2013} $$\\nThis contradicts the assumption that $R_k$ is not divisible by 2013.\\n\\nCase 2: The non-divisible integer is a column, say $C_k$ for some $k \\\\in \\\\{1, \\\\dots, 100\\\\}$.\\nThen $R_i \\\\equiv 0 \\\\pmod{2013}$ for all $i$, and $C_j \\\\equiv 0 \\\\pmod{2013}$ for all $j \\\\neq k$.\\nSubstituting these into the congruence relation:\\n$$ \\\\sum_{i=1}^{100} R_i w_i \\\\equiv 0 \\\\pmod{2013} $$\\n$$ \\\\sum_{j=1}^{100} C_j w_j \\\\equiv C_k w_k + \\\\sum_{j \\\\neq k} 0 \\\\cdot w_j \\\\equiv C_k w_k \\\\pmod{2013} $$\\nEquating these gives:\\n$$ C_k w_k \\\\equiv 0 \\\\pmod{2013} $$\\nAs before, since $\\\\gcd(w_k, 2013) = 1$, it follows that $C_k \\\\equiv 0 \\\\pmod{2013}$.\\nThis contradicts the assumption that $C_k$ is not divisible by 2013.\\n\\nSince both cases lead to a contradiction, our initial assumption that there is exactly one non-divisible integer must be false.\\nThus, there are either 0 or fewer than 1 non-divisible integers. Given that \\\"at least 199\\\" are divisible, it is impossible to have fewer than 199 divisible. If exactly 1 failed, we have a contradiction. Thus, 0 fail.\\nTherefore, all 200 integers $R_1, \\\\dots, R_{100}$ and $C_1, \\\\dots, C_{100}$ are divisible by 2013.\\n\\n## Attempt 3\\nThe problem asks us to prove that if in a $100 \\\\times 100$ table of nonzero digits, at least 199 of the 200 integers formed by rows and columns are divisible by 2013, then all 200 integers are divisible by 2013.\\n\\nLet the digits in the table be $a_{i,j}$ for $1 \\\\le i, j \\\\le 100$, where $a_{i,j} \\\\in \\\\{1, \\\\dots, 9\\\\}$.\\nLet $R_i$ be the integer represented by the $i$-th row, and $C_j$ be the integer represented by the $j$-th column.\\n$$R_i = \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j}, \\\\quad C_j = \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i}.$$\\nThe set of 200 integers is $S = \\\\{R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}\\\\}$.\\nWe are given that at least 199 elements of $S$ are divisible by 2013. This implies that at most 1 element of $S$ is not divisible by 2013.\\n\\nFirst, we factorize 2013:\\n$$2013 = 3 \\\\times 11 \\\\times 61.$$\\nSince 3, 11, and 61 are pairwise coprime, an integer is divisible by 2013 if and only if it is divisible by 3, 11, and 61. Thus, it suffices to show that under the given condition, every integer in $S$ is divisible by 3, by 11, and by 61.\\nLet $p$ be a prime factor of 2013. We will show that if at most 1 element of $S$ is not divisible by $p$, then all elements of $S$ are divisible by $p$.\\nThe prime factors are 3, 11, and 61. Notice that none of these primes divide 10, so $10$ is invertible modulo $p$.\\nLet $x_k = 10^{100-k} \\\\pmod p$ for $k=1, \\\\dots, 100$. Since $p \\\\nmid 10$, $x_k \\\\not\\\\equiv 0 \\\\pmod p$ for all $k$.\\nWe can write $R_i$ and $C_j$ as:\\n$$R_i \\\\equiv \\\\sum_{j=1}^{100} a_{i,j} x_j \\\\pmod p,$$\\n$$C_j \\\\equiv \\\\sum_{i=1}^{100} a_{i,j} x_i \\\\pmod p.$$\\n(Note: We use the same sequence $x$ for both row and column coefficients because $10^{100-j}$ and $10^{100-i}$ are just indexed terms of the sequence $10^k$).\\n\\nConsider the sum $W = \\\\sum_{i=1}^{100} x_i R_i \\\\pmod p$. Substituting the expressions for $R_i$:\\n$$W \\\\equiv \\\\sum_{i=1}^{100} x_i \\\\left( \\\\sum_{j=1}^{100} a_{i,j} x_j \\\\right) \\\\pmod p.$$\\nBy changing the order of summation:\\n$$W \\\\equiv \\\\sum_{j=1}^{100} x_j \\\\left( \\\\sum_{i=1}^{100} a_{i,j} x_i \\\\right) \\\\pmod p.$$\\nThe inner sum $\\\\sum_{i=1}^{100} a_{i,j} x_i$ is exactly $C_j$. Thus:\\n$$W \\\\equiv \\\\sum_{j=1}^{100} x_j C_j \\\\pmod p.$$\\nSo we have the identity:\\n$$\\\\sum_{i=1}^{100} x_i R_i \\\\equiv \\\\sum_{j=1}^{100} x_j C_j \\\\pmod p.$$\\n\\nNow we use the hypothesis that at most 1 number in $S$ is not divisible by $p$.\\nSuppose there is exactly one number in $S$ that is not divisible by $p$.\\nThere are two cases:\\nCase 1: The non-divisible number is a row $R_k$.\\nThen all $R_i$ for $i \\\\ne k$ are divisible by $p$, and all $C_j$ for all $j$ are divisible by $p$.\\nSubstituting into the identity:\\n$$\\\\sum_{i=1}^{100} x_i R_i \\\\equiv x_k R_k + \\\\sum_{i \\\\ne k} x_i \\\\cdot 0 \\\\equiv x_k R_k \\\\pmod p.$$\\n$$\\\\sum_{j=1}^{100} x_j C_j \\\\equiv \\\\sum_{j=1}^{100} x_j \\\\cdot 0 \\\\equiv 0 \\\\pmod p.$$\\nFrom the identity, $x_k R_k \\\\equiv 0 \\\\pmod p$.\\nSince $x_k = 10^{100-k}$ is not divisible by $p$ (as $p \\\\nmid 10$), we must have $R_k \\\\equiv 0 \\\\pmod p$. This contradicts the assumption that $R_k$ is not divisible by $p$.\\n\\nCase 2: The non-divisible number is a column $C_l$.\\nThen all $R_i$ for all $i$ are divisible by $p$, and all $C_j$ for $j \\\\ne l$ are divisible by $p$.\\nSubstituting into the identity:\\n$$\\\\sum_{i=1}^{100} x_i R_i \\\\equiv \\\\sum_{i=1}^{100} x_i \\\\cdot 0 \\\\equiv 0 \\\\pmod p.$$\\n$$\\\\sum_{j=1}^{100} x_j C_j \\\\equiv x_l C_l + \\\\sum_{j \\\\ne l} x_j \\\\cdot 0 \\\\equiv x_l C_l \\\\pmod p.$$\\nFrom the identity, $x_l C_l \\\\equiv 0 \\\\pmod p$.\\nAgain, since $x_l \\\\not\\\\equiv 0 \\\\pmod p$, we must have $C_l \\\\equiv 0 \\\\pmod p$. This contradicts the assumption that $C_l$ is not divisible by $p$.\\n\\nThus, in both cases, the assumption that a number is not divisible by $p$ leads to a contradiction. Therefore, all numbers in $S$ must be divisible by $p$.\\nThis holds for each prime factor $p \\\\in \\\\{3, 11, 61\\\\}$. Since the numbers are divisible by each of 3, 11, and 61, they are divisible by $3 \\\\times 11 \\\\times 61 = 2013$.\\nTherefore, all 200 integers are divisible by 2013.\\n\\n**Final Answer:**\\nThe problem asks to prove that all 200 numbers are divisible by 2013.\\nLet $S = \\\\{R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}\\\\}$ be the set of the 200 integers. We are given that at least 199 elements of $S$ are divisible by 2013. This means at most 1 element is not divisible by 2013.\\nSince $2013 = 3 \\\\times 11 \\\\times 61$, we consider each prime factor $p \\\\in \\\\{3, 11, 61\\\\}$ separately. For any such $p$, if $x \\\\equiv 0 \\\\pmod{2013}$, then $x \\\\equiv 0 \\\\pmod p$. Thus, at most 1 element of $S$ is not divisible by $p$.\\nLet $x_k = 10^{100-k} \\\\pmod p$. Since $p \\\\nmid 10$, $x_k \\\\not\\\\equiv 0 \\\\pmod p$ for all $k$.\\nWe observe the following congruence relation modulo $p$:\\n$$ \\\\sum_{i=1}^{100} x_i R_i = \\\\sum_{i=1}^{100} x_i \\\\sum_{j=1}^{100} a_{i,j} x_j = \\\\sum_{j=1}^{100} x_j \\\\sum_{i=1}^{100} a_{i,j} x_i = \\\\sum_{j=1}^{100} x_j C_j. $$\\nIf there were an element not divisible by $p$, say $R_k$, then all $C_j$ and $R_i$ ($i \\\\ne k$) are divisible by $p$. The relation implies $x_k R_k \\\\equiv \\\\sum x_j C_j \\\\equiv 0 \\\\pmod p$, so $R_k \\\\equiv 0 \\\\pmod p$, a contradiction. Similarly, if $C_l$ were the only non-divisible element, all $R_i$ and $C_j$ ($j \\\\ne l$) are divisible by $p$. The relation implies $x_l C_l \\\\equiv \\\\sum x_i R_i \\\\equiv 0 \\\\pmod p$, so $C_l \\\\equiv 0 \\\\pmod p$, a contradiction.\\nTherefore, no element can be not divisible by $p$. Thus, all 200 elements are divisible by 3, 11, and 61. Being divisible by these coprime factors, all 200 elements are divisible by $3 \\\\times 11 \\\\times 61 = 2013$.\\n\\nThe final result is that all of them are divisible by 2013.\\n\\n## Attempt 4\\nLet $N = 2013$. We are given a $100 \\\\times 100$ table of nonzero digits $d_{i,j} \\\\in \\\\{1, \\\\dots, 9\\\\}$ for $1 \\\\le i, j \\\\le 100$.\\nLet $R_i$ denote the integer formed by the $i$-th row and $C_j$ denote the integer formed by the $j$-th column. Specifically, the digits are concatenated from left-to-right for rows and top-to-bottom for columns. Thus:\\n$$R_i = \\\\sum_{j=1}^{100} d_{i,j} 10^{100-j}$$\\n$$C_j = \\\\sum_{i=1}^{100} d_{i,j} 10^{100-i}$$\\nWe are given that at least 199 of these 200 integers are divisible by $N$. We aim to prove that all 200 integers are divisible by $N$.\\n\\nFirst, observe that $\\\\gcd(10, 2013) = 1$. Thus, any power of 10, denoted $\\\\omega_k = 10^{100-k}$, is coprime to $2013$ and has a multiplicative inverse modulo $2013$.\\nThe problem states that for at least 199 indices pairs $(i,j)$ associated with the 200 integers, the value is $0 \\\\pmod{2013}$. Let $\\\\mathcal{I}$ be the set of indices of the integers that are divisible by $N$. We have $|\\\\mathcal{I}| \\\\ge 199$.\\nThe set of all integers consists of $\\\\{R_1, \\\\dots, R_{100}\\\\} \\\\cup \\\\{C_1, \\\\dots, C_{100}\\\\}$.\\nThe number of non-divisible integers is $200 - |\\\\mathcal{I}| \\\\le 1$.\\nLet $n_r$ be the number of rows not divisible by $N$ and $n_c$ be the number of columns not divisible by $N$. The total number of non-divisible integers is $n_r + n_c \\\\le 1$.\\nThis leaves two possibilities to consider (excluding the trivial case where 0 are non-divisible, in which case the theorem holds):\\n1. Exactly one row fails to be divisible ($n_r = 1$) and all columns are divisible ($n_c = 0$).\\n2. Exactly one column fails to be divisible ($n_c = 1$) and all rows are divisible ($n_r = 0$).\\n\\nBy symmetry (the transpose of the table swaps rows and columns but preserves the structure and divisibility properties), we only need to disprove case 1. If case 1 leads to a contradiction, case 2 leads to the same contradiction by relabeling rows and columns.\\n\\n**Disproving Case 1:**\\nAssume that there exists an index $k \\\\in \\\\{1, \\\\dots, 100\\\\}$ such that $R_k$ is not divisible by $N$, while for all $i \\\\neq k$, $R_i$ is divisible by $N$. Furthermore, assume that all columns $C_j$ are divisible by $N$.\\n\\nThe condition that all columns are divisible means that for every $j \\\\in \\\\{1, \\\\dots, 100\\\\}$:\\n$$C_j = \\\\sum_{i=1}^{100} d_{i,j} 10^{100-i} \\\\equiv 0 \\\\pmod{2013}$$\\nLet us denote the weight associated with the $m$-th position in a number as $\\\\omega_m = 10^{100-m}$. Then the conditions are:\\n1. For all $j$, $\\\\sum_{i=1}^{100} d_{i,j} \\\\omega_i \\\\equiv 0 \\\\pmod{2013}$.\\n2. For all $i \\\\neq k$, $\\\\sum_{j=1}^{100} d_{i,j} \\\\omega_j \\\\equiv 0 \\\\pmod{2013}$.\\n\\nConsider the equation for column $j$ (from condition 1):\\n$$d_{k,j} \\\\omega_k + \\\\sum_{i \\\\neq k} d_{i,j} \\\\omega_i \\\\equiv 0 \\\\pmod{2013}$$\\nSince $\\\\omega_k$ is a unit modulo $2013$ (as $\\\\gcd(10, 2013)=1$), we can isolate $d_{k,j}$:\\n$$d_{k,j} \\\\equiv -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} d_{i,j} \\\\omega_i \\\\pmod{2013}$$\\n\\nNow, consider the expression for $R_k$ (modulo $N$). By condition 2 (applied to $R_k$ itself, though we only know it's non-divisible, we can still manipulate the expression to check its value):\\n$$R_k = \\\\sum_{j=1}^{100} d_{k,j} \\\\omega_j$$\\nSubstitute the expression for $d_{k,j}$ into this sum:\\n$$R_k \\\\equiv \\\\sum_{j=1}^{100} \\\\left( -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} d_{i,j} \\\\omega_i \\\\right) \\\\omega_j \\\\pmod{2013}$$\\nRearranging the summation order:\\n$$R_k \\\\equiv -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} \\\\omega_i \\\\left( \\\\sum_{j=1}^{100} d_{i,j} \\\\omega_j \\\\right) \\\\pmod{2013}$$\\nThe inner sum $\\\\sum_{j=1}^{100} d_{i,j} \\\\omega_j$ is precisely the integer value of row $i$, denoted $R_i$.\\nThus, we have:\\n$$R_k \\\\equiv -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} \\\\omega_i R_i \\\\pmod{2013}$$\\nUnder our assumption for Case 1, $R_i$ is divisible by $N$ for all $i \\\\neq k$. Therefore, $R_i \\\\equiv 0 \\\\pmod{2013}$.\\nSubstituting $R_i \\\\equiv 0$ into the equation:\\n$$R_k \\\\equiv -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} \\\\omega_i \\\\cdot 0 \\\\equiv 0 \\\\pmod{2013}$$\\nThis implies that $R_k$ is divisible by $2013$. This contradicts the assumption that $R_k$ is not divisible by $2013$.\\n\\nTherefore, it is impossible for exactly one row to fail while all columns succeed. By symmetry, it is also impossible for exactly one column to fail while all rows succeed.\\n\\nConsequently, the only way to satisfy the condition that at least 199 of the 200 integers are divisible by 2013 is if none of them fail. That is, $n_r = 0$ and $n_c = 0$.\\n\\nThus, all 200 integers are divisible by 2013.\\n\\n**Proof that no other answer works (rigorous summary):**\\nWe established that $n_r + n_c \\\\le 1$ is required by the problem statement.\\n1. If $n_r + n_c = 0$, the condition is met trivially.\\n2. If $n_r + n_c = 1$, we proved it leads to a contradiction ($R_k \\\\equiv 0 \\\\pmod{2013}$ if assumed non-zero, or $C_m \\\\equiv 0 \\\\pmod{2013}$ if assumed non-zero).\\nSince the case $n_r + n_c > 1$ yields fewer than 199 divisible numbers (specifically, at most 198), it is excluded by the problem hypothesis.\\nThus, the only valid state is where all 200 integers are divisible.\\n\\nQ.E.D.\\n\\nThe final result is that all 200 numbers must be divisible by 2013.\\n\\nFinal Answer: All 200 of the integers are divisible by 2013.\\n\\n# Fixed response prefix\\nLet the $100 \\\\times 100$ table be denoted by $A = (a_{i,j})$ where $1 \\\\le i, j \\\\le 100$.\\nThe digits are nonzero, so $a_{i,j} \\\\in \\\\{1, 2, \\\\dots, 9\\\\}$. Wait, the problem says \\\"nonzero digits\\\".\\nThis means $a_{i,j} \\\\in \\\\{1, \\\\dots, 9\\\\}$? Or could it be that they are just integers $1, \\\\dots, 9$? \\\"Digits\\\" usually implies base 10 digits. Nonzero means $\\\\{1, \\\\dots, 9\\\\}$.\\n\\nLet $R_i$ be the integer formed by the $i$-th row for $1 \\\\le i \\\\le 100$.\\nLet $C_j$ be the integer formed by the $j$-th column for $1 \\\\le j \\\\le 100$.\\nWe have a set of $N = 200$ integers $\\\\{R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}\\\\}$.\\nLet $S$ be this set (as a multiset of values). The size is 200.\\nWe are given that at least 199 of these numbers are divisible by $2013$.\\nWe need to prove that all of them are divisible by $2013$.\\nThis is equivalent to: If $| \\\\{ x \\\\in S : x \\\\equiv 0 \\\\pmod{2013} \\\\} | \\\\ge 199$, then for all $x \\\\in S$, $x \\\\equiv 0 \\\\pmod{2013}$.\\n\\nLet's check the modulus. $2013 = 3 \\\\times 11 \\\\times 61$.\\nWait, $2+0+1+3 = 6$, so divisible by 3. $2013 / 3 = 671$.\\n$671 = 11 \\\\times 61$. So $2013 = 3 \\\\times 11 \\\\times 61$.\\nThis factorization might not be crucial immediately, but good to know.\\n\\nLet's denote $r(i)$ as the remainder of $R_i$ modulo 2013, and $c(j)$ as the remainder of $C_j$ modulo 2013.\\nThere are $m=100$ rows and $n=100$ columns.\\nLet $r_i \\\\equiv R_i \\\\pmod m$ and $c_j \\\\equiv C_j \\\\pmod m$. Actually the modulus is $M=2013$.\\nSo $R_i = \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j}$ and $C_j = \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\\nThe condition is that $r_i = 0$ or $c_j = 0$ for almost all indices.\\nSuppose exactly one number is not divisible by 2013. Without loss of generality, assume $R_k$ is not divisible, and all other $R_i$ ($i \\\\ne k$) and all $C_j$ ($1 \\\\le j \\\\le 100$) are divisible.\\nOr maybe two are not divisible, etc. But we want to show that the assumption of $\\\\ge 199$ implies all are divisible. So let's try to construct a contradiction if there is one non-divisible number.\\nOr rather, let's analyze the sum of remainders or relations between $R_i$'s and $C_j$'s.\\nSpecifically, $R_i$ depends on the row entries, $C_j$ depends on column entries. They intersect at $(i,j)$.\\n\\n# Output\\nRespond with exactly one lowercase word and nothing else -- no punctuation or\\nexplanation:\\n- For an incomplete prefix, use `yes` with probability equal to the expected\\n normalized rubric credit of a fresh continuation.\\n- For a complete response, use `yes` with probability equal to its realized\\n normalized rubric credit.\\n- Otherwise use `no`.\\n\"\n },\n {\n \"role\": \"assistant\",\n \"content\": \"yes\"\n }\n]","isConversational":true},"enable_thinking":{"kind":"bool","value":false,"string":"false"},"id":{"kind":"string","value":"8cf1cb3370041166"},"source":{"kind":"string","value":"fineproofs"},"sample_index":{"kind":"number","value":7,"string":"7"},"step_idx":{"kind":"number","value":4,"string":"4"},"num_steps":{"kind":"number","value":80,"string":"80"},"label":{"kind":"string","value":"yes"},"correct":{"kind":"bool","value":true,"string":"true"},"loss_weight":{"kind":"number","value":1,"string":"1"},"reward":{"kind":"number","value":0.9151785714285714,"string":"0.915179"}}},{"rowIdx":99,"cells":{"messages":{"kind":"list like","value":[{"role":"user","content":"# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefix is incomplete, imagine starting exactly where it ends and sampling\none fresh continuation from the same fixed policy and sampling procedure used to\ncollect the rollout data. The completed response will be graded like an olympiad\nproof: partial credit is awarded only for correct, rigorous, and clearly justified\nprogress, and the awarded points are normalized by the maximum available points.\nTreat your probability of `yes` as the expected normalized rubric credit in\nthe interval from 0 to 1.\n\nIf the prefix is a complete response, judge its realized outcome as written. Treat\nyour probability of `yes` as that response's realized normalized rubric credit in\nthe interval from 0 to 1. Do not imagine replacing or continuing an already\ncomplete response.\n\nThis is an eventual rubric-value forecast, not merely a check that the reasoning\nshown so far looks plausible. A valid prefix can still have low value when the\nremaining proof is difficult. A flawed prefix can have nonzero value when the\npolicy can recover. A guessed final answer without a valid proof earns little\ncredit unless the grading scheme says otherwise.\n\nReference attempts, when present, are side information for estimating the policy\nand problem. They are not part of the fixed prefix and are not visible to the\nhypothetical continuation.\n\n# Problem\n$10000$ nonzero digits are written in a $100$ -by- $100$ table, one digit per cell. From left to right, each row forms a $100$ -digit integer. From top to bottom, each column forms a $100$ -digit integer. So the rows and columns form $200$ integers (each with $100$ digits), not necessarily distinct. Prove that if at least $199$ of these $200$ numbers are divisible by $2013$ , then all of them are divisible by $2013$ .\n\n# Reference attempts\n## Attempt 1\nLet $N = 2013$. We are given a $100 \\times 100$ table of nonzero digits $d_{i,j} \\in \\{1, \\dots, 9\\}$ for $1 \\le i, j \\le 100$.\nLet $R_i$ denote the integer formed by the $i$-th row and $C_j$ denote the integer formed by the $j$-th column. Specifically, the digits are concatenated from left-to-right for rows and top-to-bottom for columns. Thus:\n$$R_i = \\sum_{j=1}^{100} d_{i,j} 10^{100-j}$$\n$$C_j = \\sum_{i=1}^{100} d_{i,j} 10^{100-i}$$\nWe are given that at least 199 of these 200 integers are divisible by $N$. We aim to prove that all 200 integers are divisible by $N$.\n\nFirst, observe that $\\gcd(10, 2013) = 1$. Thus, any power of 10, denoted $\\omega_k = 10^{100-k}$, is coprime to $2013$ and has a multiplicative inverse modulo $2013$.\nThe problem states that for at least 199 indices pairs $(i,j)$ associated with the 200 integers, the value is $0 \\pmod{2013}$. Let $\\mathcal{I}$ be the set of indices of the integers that are divisible by $N$. We have $|\\mathcal{I}| \\ge 199$.\nThe set of all integers consists of $\\{R_1, \\dots, R_{100}\\} \\cup \\{C_1, \\dots, C_{100}\\}$.\nThe number of non-divisible integers is $200 - |\\mathcal{I}| \\le 1$.\nLet $n_r$ be the number of rows not divisible by $N$ and $n_c$ be the number of columns not divisible by $N$. The total number of non-divisible integers is $n_r + n_c \\le 1$.\nThis leaves two possibilities to consider (excluding the trivial case where 0 are non-divisible, in which case the theorem holds):\n1. Exactly one row fails to be divisible ($n_r = 1$) and all columns are divisible ($n_c = 0$).\n2. Exactly one column fails to be divisible ($n_c = 1$) and all rows are divisible ($n_r = 0$).\n\nBy symmetry (the transpose of the table swaps rows and columns but preserves the structure and divisibility properties), we only need to disprove case 1. If case 1 leads to a contradiction, case 2 leads to the same contradiction by relabeling rows and columns.\n\n**Disproving Case 1:**\nAssume that there exists an index $k \\in \\{1, \\dots, 100\\}$ such that $R_k$ is not divisible by $N$, while for all $i \\neq k$, $R_i$ is divisible by $N$. Furthermore, assume that all columns $C_j$ are divisible by $N$.\n\nThe condition that all columns are divisible means that for every $j \\in \\{1, \\dots, 100\\}$:\n$$C_j = \\sum_{i=1}^{100} d_{i,j} 10^{100-i} \\equiv 0 \\pmod{2013}$$\nLet us denote the weight associated with the $m$-th position in a number as $\\omega_m = 10^{100-m}$. Then the conditions are:\n1. For all $j$, $\\sum_{i=1}^{100} d_{i,j} \\omega_i \\equiv 0 \\pmod{2013}$.\n2. For all $i \\neq k$, $\\sum_{j=1}^{100} d_{i,j} \\omega_j \\equiv 0 \\pmod{2013}$.\n\nConsider the equation for column $j$ (from condition 1):\n$$d_{k,j} \\omega_k + \\sum_{i \\neq k} d_{i,j} \\omega_i \\equiv 0 \\pmod{2013}$$\nSince $\\omega_k$ is a unit modulo $2013$ (as $\\gcd(10, 2013)=1$), we can isolate $d_{k,j}$:\n$$d_{k,j} \\equiv -\\omega_k^{-1} \\sum_{i \\neq k} d_{i,j} \\omega_i \\pmod{2013}$$\n\nNow, consider the expression for $R_k$ (modulo $N$). By condition 2 (applied to $R_k$ itself, though we only know it's non-divisible, we can still manipulate the expression to check its value):\n$$R_k = \\sum_{j=1}^{100} d_{k,j} \\omega_j$$\nSubstitute the expression for $d_{k,j}$ into this sum:\n$$R_k \\equiv \\sum_{j=1}^{100} \\left( -\\omega_k^{-1} \\sum_{i \\neq k} d_{i,j} \\omega_i \\right) \\omega_j \\pmod{2013}$$\nRearranging the summation order:\n$$R_k \\equiv -\\omega_k^{-1} \\sum_{i \\neq k} \\omega_i \\left( \\sum_{j=1}^{100} d_{i,j} \\omega_j \\right) \\pmod{2013}$$\nThe inner sum $\\sum_{j=1}^{100} d_{i,j} \\omega_j$ is precisely the integer value of row $i$, denoted $R_i$.\nThus, we have:\n$$R_k \\equiv -\\omega_k^{-1} \\sum_{i \\neq k} \\omega_i R_i \\pmod{2013}$$\nUnder our assumption for Case 1, $R_i$ is divisible by $N$ for all $i \\neq k$. Therefore, $R_i \\equiv 0 \\pmod{2013}$.\nSubstituting $R_i \\equiv 0$ into the equation:\n$$R_k \\equiv -\\omega_k^{-1} \\sum_{i \\neq k} \\omega_i \\cdot 0 \\equiv 0 \\pmod{2013}$$\nThis implies that $R_k$ is divisible by $2013$. This contradicts the assumption that $R_k$ is not divisible by $2013$.\n\nTherefore, it is impossible for exactly one row to fail while all columns succeed. By symmetry, it is also impossible for exactly one column to fail while all rows succeed.\n\nConsequently, the only way to satisfy the condition that at least 199 of the 200 integers are divisible by 2013 is if none of them fail. That is, $n_r = 0$ and $n_c = 0$.\n\nThus, all 200 integers are divisible by 2013.\n\n**Proof that no other answer works (rigorous summary):**\nWe established that $n_r + n_c \\le 1$ is required by the problem statement.\n1. If $n_r + n_c = 0$, the condition is met trivially.\n2. If $n_r + n_c = 1$, we proved it leads to a contradiction ($R_k \\equiv 0 \\pmod{2013}$ if assumed non-zero, or $C_m \\equiv 0 \\pmod{2013}$ if assumed non-zero).\nSince the case $n_r + n_c > 1$ yields fewer than 199 divisible numbers (specifically, at most 198), it is excluded by the problem hypothesis.\nThus, the only valid state is where all 200 integers are divisible.\n\nQ.E.D.\n\nThe final result is that all 200 numbers must be divisible by 2013.\n\nFinal Answer: All 200 of the integers are divisible by 2013.\n\n## Attempt 2\nLet $A = (a_{i,j})$ be the $100 \\times 100$ table of nonzero digits, where $1 \\le i, j \\le 100$.\nFor each row $i$, let $R_i$ be the integer formed by the digits $a_{i,1}a_{i,2}\\dots a_{i,100}$.\nThe value is given by $R_i = \\sum_{j=1}^{100} a_{i,j} 10^{100-j}$.\nFor each column $j$, let $C_j$ be the integer formed by the digits $a_{1,j}a_{2,j}\\dots a_{100,j}$.\nThe value is given by $C_j = \\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\nLet $M = 2013$. We are given that at least 199 of the integers $R_1, \\dots, R_{100}, C_1, \\dots, C_{100}$ are divisible by $M$.\nWe must prove that all 200 integers are divisible by $M$.\n\nSuppose, for the sake of contradiction, that at least one integer is not divisible by $M$. Since at most 1 is not divisible (as 199 are divisible), there exists exactly one integer among the 200 that is not divisible by $M$.\n\nLet us define weights $w_k = 10^{100-k}$ for $k=1, \\dots, 100$.\nThen $R_i = \\sum_{j=1}^{100} a_{i,j} w_j$ and $C_j = \\sum_{i=1}^{100} a_{i,j} w_i$.\nConsider the scalar quantity $S = \\sum_{i=1}^{100} \\sum_{j=1}^{100} a_{i,j} w_i w_j$.\nWe can express $S$ in two ways by grouping the terms:\n$$S = \\sum_{j=1}^{100} w_j \\left( \\sum_{i=1}^{100} a_{i,j} w_i \\right) = \\sum_{j=1}^{100} w_j C_j$$\n$$S = \\sum_{i=1}^{100} w_i \\left( \\sum_{j=1}^{100} a_{i,j} w_j \\right) = \\sum_{i=1}^{100} w_i R_i$$\n\nCase 1: Suppose the non-divisible integer is a column $C_{k_0}$ (with $1 \\le k_0 \\le 100$).\nThen $C_j \\equiv 0 \\pmod M$ for all $j \\ne k_0$, and $C_{k_0} \\not\\equiv 0 \\pmod M$.\nAlso, all rows are divisible, so $R_i \\equiv 0 \\pmod M$ for all $i$.\nUsing the expression $S = \\sum_{i=1}^{100} w_i R_i$, we substitute $R_i \\equiv 0$:\n$$S \\equiv \\sum_{i=1}^{100} w_i \\cdot 0 \\equiv 0 \\pmod M.$$\nUsing the expression $S = \\sum_{j=1}^{100} w_j C_j$, we substitute $C_j \\equiv 0$ for $j \\ne k_0$:\n$$S \\equiv \\sum_{j \\ne k_0} w_j \\cdot 0 + w_{k_0} C_{k_0} \\equiv w_{k_0} C_{k_0} \\pmod M.$$\nCombining these, we get $w_{k_0} C_{k_0} \\equiv 0 \\pmod M$.\nNote that $w_{k_0} = 10^{100-k_0}$. Since $\\gcd(10, M) = \\gcd(10, 3 \\cdot 11 \\cdot 61) = 1$, the weight $w_{k_0}$ is coprime to $M$ and thus is invertible modulo $M$.\nTherefore, $w_{k_0} C_{k_0} \\equiv 0 \\implies C_{k_0} \\equiv 0 \\pmod M$, which contradicts the assumption that $C_{k_0}$ is not divisible by $M$.\n\nCase 2: Suppose the non-divisible integer is a row $R_{i_0}$ (with $1 \\le i_0 \\le 100$).\nThen $R_i \\equiv 0 \\pmod M$ for all $i \\ne i_0$, and $R_{i_0} \\not\\equiv 0 \\pmod M$.\nAlso, all columns are divisible, so $C_j \\equiv 0 \\pmod M$ for all $j$.\nUsing $S = \\sum_{i=1}^{100} w_i R_i$, we substitute $R_i \\equiv 0$ for $i \\ne i_0$:\n$$S \\equiv w_{i_0} R_{i_0} + \\sum_{i \\ne i_0} w_i \\cdot 0 \\equiv w_{i_0} R_{i_0} \\pmod M.$$\nUsing $S = \\sum_{j=1}^{100} w_j C_j$, we substitute $C_j \\equiv 0$ for all $j$:\n$$S \\equiv \\sum_{j=1}^{100} w_j \\cdot 0 \\equiv 0 \\pmod M.$$\nThis implies $w_{i_0} R_{i_0} \\equiv 0 \\pmod M$. Again, $w_{i_0}$ is invertible modulo $M$, so $R_{i_0} \\equiv 0 \\pmod M$, contradicting the assumption that $R_{i_0}$ is not divisible.\n\nThus, in either case, the assumption that exactly one integer is not divisible leads to a contradiction.\nTherefore, it is impossible for exactly 199 integers to be divisible by 2013. Since at least 199 are divisible, all 200 must be divisible by 2013.\n\nFinal Result:\nAll 200 integers formed by the rows and columns are divisible by 2013.\n\n## Attempt 3\nLet $n = 100$ and $M = 2013$.\nLet $d_{i,j}$ be the digit in the $i$-th row and $j$-th column, where $1 \\le i, j \\le n$. The problem states that $d_{i,j} \\in \\{1, 2, \\dots, 9\\}$.\nThe integer formed by the $i$-th row is denoted by $R_i$. Based on the convention of writing digits from left to right, $R_i$ is given by:\n$$ R_i = \\sum_{j=1}^{n} d_{i,j} 10^{n-j} $$\nSimilarly, the integer formed by the $j$-th column is denoted by $C_j$:\n$$ C_j = \\sum_{i=1}^{n} d_{i,j} 10^{n-i} $$\nWe are given that at least $2n-1$ of these $2n$ integers ($R_1, \\dots, R_n, C_1, \\dots, C_n$) are divisible by $M$. We must prove that all of them are divisible by $M$.\n\nConsider the following algebraic identity involving the row and column sums:\n$$ S = \\sum_{i=1}^{n} R_i 10^{n-i} $$\nSubstituting the expression for $R_i$:\n$$ S = \\sum_{i=1}^{n} 10^{n-i} \\left( \\sum_{j=1}^{n} d_{i,j} 10^{n-j} \\right) = \\sum_{i=1}^{n} \\sum_{j=1}^{n} d_{i,j} 10^{2n - i - j} $$\nSimilarly, consider the sum of the column integers weighted by powers of 10:\n$$ T = \\sum_{j=1}^{n} C_j 10^{n-j} $$\nSubstituting the expression for $C_j$:\n$$ T = \\sum_{j=1}^{n} 10^{n-j} \\left( \\sum_{i=1}^{n} d_{i,j} 10^{n-i} \\right) = \\sum_{j=1}^{n} \\sum_{i=1}^{n} d_{i,j} 10^{2n - i - j} $$\nSince the summation variables $i$ and $j$ run over the same range $[1, n]$, the double sums are identical:\n$$ \\sum_{i=1}^{n} \\sum_{j=1}^{n} d_{i,j} 10^{2n - i - j} = \\sum_{j=1}^{n} \\sum_{i=1}^{n} d_{i,j} 10^{2n - i - j} $$\nThus, we have the identity:\n$$ \\sum_{i=1}^{n} R_i 10^{n-i} \\equiv \\sum_{j=1}^{n} C_j 10^{n-j} \\pmod M $$\nThis identity holds as an equality of integers, and consequently as a congruence modulo $M$.\n\nLet $S_R = \\{ i \\in \\{1, \\dots, n\\} \\mid R_i \\equiv 0 \\pmod M \\}$ and $S_C = \\{ j \\in \\{1, \\dots, n\\} \\mid C_j \\equiv 0 \\pmod M \\}$.\nThe condition that at least $199$ of the $200$ numbers are divisible by $M$ implies that $|S_R| + |S_C| \\ge 199$.\nSince $|S_R| \\le 100$ and $|S_C| \\le 100$, the possible values for $(|S_R|, |S_C|)$ are limited to cases where the number of non-divisible integers is at most 1 (since $200 - 199 = 1$). Specifically, the number of indices not in $S_R \\cup S_C$ is $k$ such that $200 - |S_R| - |S_C| \\ge 200 - 199 = 1$, so the count of non-divisible numbers is $N_{bad} = 200 - |S_R| - |S_C| \\le 1$.\n\nWe examine the possible scenarios for the set of \"bad\" (non-divisible) numbers:\n1. **Zero bad numbers:** $|S_R| = 100$ and $|S_C| = 100$. In this case, all 200 numbers are divisible by $M$, and we are done.\n2. **One bad number:** There are two subcases due to the symmetry between rows and columns.\n * *Subcase 2a:* $R_i \\equiv 0 \\pmod M$ for all $i$, but $C_n \\not\\equiv 0 \\pmod M$ (so $|S_R|=100, |S_C|=99$).\n Applying the identity modulo $M$:\n $$ \\sum_{i=1}^{n} R_i 10^{n-i} \\equiv \\sum_{j=1}^{n} C_j 10^{n-j} \\pmod M $$\n The Left Hand Side (LHS) is:\n $$ \\sum_{i=1}^{n} 0 \\cdot 10^{n-i} \\equiv 0 \\pmod M $$\n The Right Hand Side (RHS) is:\n $$ \\sum_{j=1}^{n-1} C_j 10^{n-j} + C_n 10^{n-n} $$\n Since $C_1, \\dots, C_{n-1}$ are divisible by $M$, the first part is $0$. The term $10^{n-n} = 10^0 = 1$.\n So RHS $\\equiv C_n \\pmod M$.\n Thus, $0 \\equiv C_n \\pmod M$, which implies $C_n$ is divisible by $M$. This contradicts the assumption that $C_n$ is bad. Therefore, this case is impossible.\n * *Subcase 2b:* $C_j \\equiv 0 \\pmod M$ for all $j$, but $R_n \\not\\equiv 0 \\pmod M$ (so $|S_R|=99, |S_C|=100$).\n By symmetry with Subcase 2a, we have:\n $$ \\sum_{i=1}^{n} R_i 10^{n-i} \\equiv \\sum_{j=1}^{n} C_j 10^{n-j} \\pmod M $$\n $$ R_n \\cdot 10^0 + 0 \\equiv 0 \\implies R_n \\equiv 0 \\pmod M $$\n Again, this is a contradiction.\n\nAny other configuration would involve at least 2 bad numbers (e.g., $R_n$ and $C_n$ both bad, or $R_a$ and $R_b$ both bad), resulting in at most $200-2 = 198$ divisible numbers, which violates the hypothesis.\n\nConsequently, the only possible configuration consistent with the hypothesis is the first one: all 200 numbers are divisible by $M$. Thus, the claim is proven.\n\n**Answer:**\nThe integers formed by the rows are $R_i = \\sum_{j=1}^{100} d_{i,j} 10^{100-j}$ and the integers formed by the columns are $C_j = \\sum_{i=1}^{100} d_{i,j} 10^{100-i}$. Consider the sums:\n$$ \\sum_{i=1}^{100} R_i 10^{100-i} \\quad \\text{and} \\quad \\sum_{j=1}^{100} C_j 10^{100-j}. $$\nExpanding these sums reveals that both are equal to $\\sum_{i=1}^{100} \\sum_{j=1}^{100} d_{i,j} 10^{200-i-j}$. Hence,\n$$ \\sum_{i=1}^{100} R_i 10^{100-i} \\equiv \\sum_{j=1}^{100} C_j 10^{100-j} \\pmod{2013}. $$\nAssume that at least 199 of the numbers are divisible by 2013. This implies that at most one number is not divisible. If there are no non-divisible numbers, we are done. If there is exactly one non-divisible number, symmetry allows us to assume it is a column integer $C_n$ (the other case is symmetric to considering $R_n$). Then $R_i \\equiv 0 \\pmod{2013}$ for all $i$ and $C_j \\equiv 0 \\pmod{2013}$ for all $j < n$. Substituting these into the identity yields:\n$$ \\sum_{i=1}^{100} 0 \\cdot 10^{100-i} \\equiv \\sum_{j=1}^{n-1} 0 \\cdot 10^{100-j} + C_n \\cdot 10^0 \\pmod{2013} $$\n$$ 0 \\equiv C_n \\pmod{2013}. $$\nThis implies $C_n$ is divisible by 2013, contradicting the assumption that it was non-divisible. Therefore, it is impossible to have exactly one non-divisible number. The hypothesis implies that all 200 numbers must be divisible by 2013.\n\n## Attempt 4\nLet the digits in the table be denoted by $a_{ij}$ for $1 \\le i, j \\le 100$, where $a_{ij} \\in \\{1, 2, \\dots, 9\\}$ are the nonzero digits.\nLet $R_i$ be the integer formed by the $i$-th row, and $C_j$ be the integer formed by the $j$-th column.\nBased on the positional notation in base 10, we can express these integers as follows:\n$$ R_i = \\sum_{j=1}^{100} a_{ij} 10^{100-j} $$\n$$ C_j = \\sum_{i=1}^{100} a_{ij} 10^{100-i} $$\nLet us define a sequence of weights $w_k = 10^{100-k}$ for $k = 1, 2, \\dots, 100$.\nThen the expressions become:\n$$ R_i = \\sum_{j=1}^{100} a_{ij} w_j $$\n$$ C_j = \\sum_{i=1}^{100} a_{ij} w_i $$\nWe consider the quantity $S = \\sum_{i=1}^{100} R_i w_i$. Substituting the expression for $R_i$, we get:\n$$ \\begin{aligned} S &= \\sum_{i=1}^{100} w_i \\left( \\sum_{j=1}^{100} a_{ij} w_j \\right) \\\\ &= \\sum_{i=1}^{100} \\sum_{j=1}^{100} a_{ij} w_i w_j \\end{aligned} $$\nSimilarly, consider the quantity $T = \\sum_{j=1}^{100} C_j w_j$. Substituting the expression for $C_j$, we get:\n$$ \\begin{aligned} T &= \\sum_{j=1}^{100} w_j \\left( \\sum_{i=1}^{100} a_{ij} w_i \\right) \\\\ &= \\sum_{j=1}^{100} \\sum_{i=1}^{100} a_{ij} w_j w_i \\end{aligned} $$\nSince finite sums commute and multiplication of integers is commutative ($w_i w_j = w_j w_i$), we observe that $S = T$. That is:\n$$ \\sum_{i=1}^{100} R_i w_i = \\sum_{j=1}^{100} C_j w_j $$\nThis equality holds over the integers. Therefore, it also holds modulo any integer $m$. In particular, let $m = 2013$.\n$$ \\sum_{i=1}^{100} R_i w_i \\equiv \\sum_{j=1}^{100} C_j w_j \\pmod{2013} $$\nWe are given that at least 199 of the 200 integers $\\{R_1, \\dots, R_{100}, C_1, \\dots, C_{100}\\}$ are divisible by 2013. This means there is at most 1 integer among them that is not divisible by 2013.\nWe proceed by contradiction. Suppose there exists exactly one integer $X$ in the set which is not divisible by 2013. There are two cases:\nCase 1: The non-divisible integer is a row, say $R_k$ for some $k \\in \\{1, \\dots, 100\\}$.\nThen $R_i \\equiv 0 \\pmod{2013}$ for all $i \\neq k$, and $C_j \\equiv 0 \\pmod{2013}$ for all $j \\in \\{1, \\dots, 100\\}$.\nSubstituting these into our congruence relation:\n$$ \\sum_{i=1}^{100} R_i w_i \\equiv R_k w_k + \\sum_{i \\neq k} 0 \\cdot w_i \\equiv R_k w_k \\pmod{2013} $$\n$$ \\sum_{j=1}^{100} C_j w_j \\equiv \\sum_{j=1}^{100} 0 \\cdot w_j \\equiv 0 \\pmod{2013} $$\nEquating these results gives:\n$$ R_k w_k \\equiv 0 \\pmod{2013} $$\nRecall that $w_k = 10^{100-k}$. The modulus is $2013$. We factorize 2013 as $2013 = 3 \\times 11 \\times 61$.\nSince $10$ is not divisible by 3, 11, or 61 (obvious inspection or calculation), $\\gcd(10, 2013) = 1$. Consequently, $\\gcd(w_k, 2013) = \\gcd(10^{100-k}, 2013) = 1$.\nThis implies that $w_k$ is invertible modulo 2013. Multiplying both sides of $R_k w_k \\equiv 0 \\pmod{2013}$ by the modular inverse of $w_k$, we obtain:\n$$ R_k \\equiv 0 \\pmod{2013} $$\nThis contradicts the assumption that $R_k$ is not divisible by 2013.\n\nCase 2: The non-divisible integer is a column, say $C_k$ for some $k \\in \\{1, \\dots, 100\\}$.\nThen $R_i \\equiv 0 \\pmod{2013}$ for all $i$, and $C_j \\equiv 0 \\pmod{2013}$ for all $j \\neq k$.\nSubstituting these into the congruence relation:\n$$ \\sum_{i=1}^{100} R_i w_i \\equiv 0 \\pmod{2013} $$\n$$ \\sum_{j=1}^{100} C_j w_j \\equiv C_k w_k + \\sum_{j \\neq k} 0 \\cdot w_j \\equiv C_k w_k \\pmod{2013} $$\nEquating these gives:\n$$ C_k w_k \\equiv 0 \\pmod{2013} $$\nAs before, since $\\gcd(w_k, 2013) = 1$, it follows that $C_k \\equiv 0 \\pmod{2013}$.\nThis contradicts the assumption that $C_k$ is not divisible by 2013.\n\nSince both cases lead to a contradiction, our initial assumption that there is exactly one non-divisible integer must be false.\nThus, there are either 0 or fewer than 1 non-divisible integers. Given that \"at least 199\" are divisible, it is impossible to have fewer than 199 divisible. If exactly 1 failed, we have a contradiction. Thus, 0 fail.\nTherefore, all 200 integers $R_1, \\dots, R_{100}$ and $C_1, \\dots, C_{100}$ are divisible by 2013.\n\n# Fixed response prefix\nLet the $100 \\times 100$ table be denoted by $A = (a_{i,j})$ where $1 \\le i, j \\le 100$.\nThe digits are nonzero, so $a_{i,j} \\in \\{1, 2, \\dots, 9\\}$. Wait, the problem says \"nonzero digits\".\nThis means $a_{i,j} \\in \\{1, \\dots, 9\\}$? Or could it be that they are just integers $1, \\dots, 9$? \"Digits\" usually implies base 10 digits. Nonzero means $\\{1, \\dots, 9\\}$.\n\nLet $R_i$ be the integer formed by the $i$-th row for $1 \\le i \\le 100$.\nLet $C_j$ be the integer formed by the $j$-th column for $1 \\le j \\le 100$.\nWe have a set of $N = 200$ integers $\\{R_1, \\dots, R_{100}, C_1, \\dots, C_{100}\\}$.\nLet $S$ be this set (as a multiset of values). The size is 200.\nWe are given that at least 199 of these numbers are divisible by $2013$.\nWe need to prove that all of them are divisible by $2013$.\nThis is equivalent to: If $| \\{ x \\in S : x \\equiv 0 \\pmod{2013} \\} | \\ge 199$, then for all $x \\in S$, $x \\equiv 0 \\pmod{2013}$.\n\nLet's check the modulus. $2013 = 3 \\times 11 \\times 61$.\nWait, $2+0+1+3 = 6$, so divisible by 3. $2013 / 3 = 671$.\n$671 = 11 \\times 61$. So $2013 = 3 \\times 11 \\times 61$.\nThis factorization might not be crucial immediately, but good to know.\n\nLet's denote $r(i)$ as the remainder of $R_i$ modulo 2013, and $c(j)$ as the remainder of $C_j$ modulo 2013.\nThere are $m=100$ rows and $n=100$ columns.\nLet $r_i \\equiv R_i \\pmod m$ and $c_j \\equiv C_j \\pmod m$. Actually the modulus is $M=2013$.\nSo $R_i = \\sum_{j=1}^{100} a_{i,j} 10^{100-j}$ and $C_j = \\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\nThe condition is that $r_i = 0$ or $c_j = 0$ for almost all indices.\nSuppose exactly one number is not divisible by 2013. Without loss of generality, assume $R_k$ is not divisible, and all other $R_i$ ($i \\ne k$) and all $C_j$ ($1 \\le j \\le 100$) are divisible.\nOr maybe two are not divisible, etc. But we want to show that the assumption of $\\ge 199$ implies all are divisible. So let's try to construct a contradiction if there is one non-divisible number.\nOr rather, let's analyze the sum of remainders or relations between $R_i$'s and $C_j$'s.\nSpecifically, $R_i$ depends on the row entries, $C_j$ depends on column entries. They intersect at $(i,j)$.\n\nLet $x_{i,j} = a_{i,j}$. These are variables in $\\{1, \\dots, 9\\}$. This constraint might be used to prevent trivial counterexamples (e.g., setting some digits to make things cancel out mod M easily? No, the range is restricted).\nHowever, $100$ digits forming an integer means $R_i = \\sum_{j=1}^{100} x_{i,j} 10^{100-j}$.\nNotice that $10^{100} \\equiv 1 \\pmod 9$. Modulo 3, $10 \\equiv 1$.\nModulo 11, $10 \\equiv -1$.\nModulo 61...\n\nLet's look at the structure of the linear map from the space of matrices to the values modulo $M$.\nLet $\\phi(M)$ be the map $(a_{i,j}) \\mapsto (\\text{values } R_i, C_j \\bmod M)$.\nThis looks like we are trying to solve a system of linear congruences over the matrix variables.\nBut the number of equations is 200, number of variables is 10000.\nActually, the divisibility condition is about the specific values.\nIs there a simpler way?\nConsider the sum of the remainders. Or sum of digits.\nIf all numbers are divisible by 3, sum of digits of each must be divisible by 3.\nFor rows, $\\sum_{j=1}^{100} a_{i,j}$ is sum of digits of $R_i$.\nSince $R_i$ is formed by digits $a_{i,j}$, $R_i \\equiv \\sum a_{i,j} \\pmod 9$.\nWait, $M=2013$ is not 9. It is divisible by 3. So $R_i \\equiv 0 \\pmod 3 \\iff \\sum a_{i,j} \\equiv 0 \\pmod 3$.\nSimilarly for columns.\nIf at least 199 integers are divisible by 2013, they are certainly divisible by 3.\nThis imposes constraints on the sums of digits in rows and columns.\nLet $s_{i} = \\sum_{j=1}^{100} a_{i,j}$ and $t_{j} = \\sum_{i=1}^{100} a_{i,j}$.\n$s_i$ is the sum of digits in row $i$. $t_j$ is sum of digits in column $j$.\n$\\sum s_i = \\sum t_j = \\sum a_{i,j}$.\nIf $R_i \\equiv 0 \\pmod 3$, then $s_i \\equiv 0 \\pmod 3$.\nIf $C_j \\equiv 0 \\pmod 3$, then $t_j \\equiv 0 \\pmod 3$.\nSuppose exactly one row $R_k$ is NOT divisible by 2013. Then all $R_i$ ($i \\ne k$) are divisible by 2013 $\\implies$ divisible by 3.\nAll $C_j$ are divisible by 2013 $\\implies$ divisible by 3.\nThen $s_i \\equiv 0 \\pmod 3$ for $i \\ne k$, and $t_j \\equiv 0 \\pmod 3$ for all $j$.\nWhat about $s_k$?\n$\\sum_{j=1}^{100} a_{k,j} = s_k$. Also $s_k = \\sum_{j} a_{k,j}$.\nFrom $t_j \\equiv 0 \\pmod 3$, we have $\\sum_{i} a_{i,j} \\equiv 0 \\pmod 3$.\nSumming over $j$: $\\sum_{j} \\sum_{i} a_{i,j} = \\sum s_i \\equiv 0 \\pmod 3$.\nWe have $s_k + \\sum_{i \\ne k} s_i \\equiv 0 \\pmod 3$.\nSince $s_i \\equiv 0 \\pmod 3$ for $i \\ne k$, this gives $s_k \\equiv 0 \\pmod 3$.\nThus $s_k \\equiv 0 \\pmod 3$.\nSo $R_k \\equiv \\sum a_{k,j} \\equiv 0 \\pmod 3$.\nThis doesn't help with divisibility by 2013 directly because $2013$ has factors 11 and 61.\nBut $R_k \\equiv 0 \\pmod 3$ is necessary. Is it possible that $R_k$ is not divisible by 2013 but is divisible by 3? Yes.\nHowever, the condition is stronger. 199 out of 200 are divisible by 2013.\nLet's see. Suppose $R_k$ is the only one not divisible.\nThen $R_k \\not\\equiv 0 \\pmod{2013}$.\nWe established that $R_k \\equiv 0 \\pmod 3$.\nSimilarly, consider modulo 11. $2013 = 3 \\cdot 11 \\cdot 61$.\nIf 199 numbers are divisible by $M$, they are divisible by 11.\nIf only one number $x$ is not divisible by $M$, it must be divisible by 11 or not?\nWait. If $x$ is not divisible by $M$, it means $x \\not\\equiv 0 \\pmod{M}$.\nThis does NOT imply $x \\not\\equiv 0 \\pmod p$ for any factor $p$.\nHowever, the hypothesis says \"at least 199 ... are divisible by 2013\".\nLet $A$ be the set of numbers. Let $Z$ be the subset of numbers divisible by 2013.\n$|Z| \\ge 199$. So the complement $Z^c$ has size $\\le 1$.\nCase 1: All numbers are divisible. Done.\nCase 2: There is exactly one number not divisible.\nWithout loss of generality, suppose it is $R_k$. Then all $R_i$ ($i \\ne k$) and all $C_j$ are divisible by 2013.\nIn particular, they are divisible by 11.\n$R_i \\equiv \\sum_{j=1}^{100} a_{i,j} (-1)^{100-j} \\equiv \\sum_{j=1}^{100} a_{i,j} (-1)^{j} \\pmod{11}$ since $(-1)^{100}=1$.\nWait, digit $d$ at position $j$ (from left, $j=1$) has weight $10^{100-j}$.\n$j=1 \\implies 10^{99} \\equiv (-1)^{99} \\equiv -1 \\pmod{11}$.\n$j=100 \\implies 10^0 = 1 \\equiv 1 \\pmod{11}$.\nSo $R_i \\equiv \\sum_{j=1}^{100} a_{i,j} (-1)^{100-j} \\pmod{11}$.\nSince $100-j$ is even for even $j$, odd for odd $j$.\nSo $R_i \\equiv \\sum_{j=1}^{100} (-1)^{j+1} a_{i,j} \\pmod{11}$ ? Let's check signs.\n$100-1=99$ odd $\\to -1$. $100-2=98$ even $\\to 1$.\nSo coeff of $a_{i,j}$ is $(-1)^{100-j}$.\nLet's denote $w_j = (-1)^{100-j}$. Note $w_j = (-1)^j w_1$. $w_1 = -1$. So $w_j = -(-1)^j = (-1)^{j+1}$.\nActually let's just write $\\sum_{j} (-1)^{j} a_{i,j}$ with adjusted sign.\nLet's use alternating sum of digits.\nLet $A_i$ be the alternating sum of row $i$. $A_i = \\sum_{j=1}^{100} (-1)^{j-1} a_{i,j}$. Or something similar.\nDivisibility by 11 depends on this.\nIf all $C_j$ are divisible by 11, then the alternating sum of each column is divisible by 11.\nIf all $R_i$ for $i \\ne k$ are divisible by 11, then their alternating sums are divisible by 11.\nDoes this force $R_k$ to be divisible by 11?\nLet's define $L(R) = \\sum_{j=1}^{100} (-1)^{j} a_{1,j} + \\dots$. No, row index $i$.\n$L_i = \\sum_{j=1}^{100} (-1)^j a_{i,j}$. (Using some fixed convention, say $100-1$ is odd power of 10).\nLet's fix the weights: $P_j = 10^{100-j}$.\n$R_i = \\sum_{j=1}^{100} P_j a_{i,j}$.\n$C_j = \\sum_{i=1}^{100} P_i a_{i,j}$ (Wait, column digits are ordered top to bottom, so position $i$ has weight $10^{100-i}$).\nYes, $C_j = \\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\nNote the weights are the same function of index. Let $W_k = 10^{100-k}$.\n$R_i = \\sum_{j=1}^{100} W_j a_{i,j}$.\n$C_j = \\sum_{i=1}^{100} W_i a_{i,j}$.\n\nWe have 100 conditions for $C_j$ being 0 mod 2013, and 100 conditions for $R_i$ being 0 mod 2013.\nActually, we have 199 conditions satisfied.\nAssume $R_k$ is the exceptional one. So $R_k \\not\\equiv 0$.\nAll $C_j \\equiv 0 \\pmod{2013}$.\nAlso $R_i \\equiv 0 \\pmod{2013}$ for all $i \\ne k$.\nConsider the determinant or something? No.\nLook at the intersection of sets of constraints.\nLet's work in the ring $\\mathbb{Z}_{2013}$.\nThe $C_j$'s give information about the weighted column sums.\nThe $R_i$'s give info about weighted row sums.\nIs it possible that we can deduce something about the structure of the matrix?\nThe digits $a_{i,j}$ are just numbers.\nMaybe we can sum $R_i$ or something.\nOr sum $C_j$.\nBut that sums everything up twice.\nConsider $\\sum_{i} R_i = \\sum_{i} \\sum_{j} W_j a_{i,j} = \\sum_{j} W_j (\\sum_{i} a_{i,j}) = \\sum_{j} W_j T_j$ where $T_j$ is sum of col $j$.\nThis doesn't seem to link $R$ and $C$ strongly.\nHowever, we know $R_i$ and $C_j$ individually.\n\nLet's look at the modular arithmetic properties.\nThe modulus is $M = 2013 = 3 \\cdot 11 \\cdot 61$.\nIt turns out that if a set of numbers are multiples of $M$, then certain algebraic relations must hold.\nPerhaps we can show that $R_k \\equiv \\text{something derived from } C_j \\text{ and } R_i$?\nWait, the $C_j$ are fully known (they are 0 mod M).\nSo $C_j \\equiv 0$ for all $j$.\nThis means $\\sum_{i=1}^{100} a_{i,j} W_i \\equiv 0 \\pmod M$.\nHere $a_{i,j}$ are unknowns.\nThe equation $C_j \\equiv 0$ holds for $j=1,\\dots,100$.\nThis is a system of 100 linear equations in $a_{i,j}$ (with coefficients $W_i$).\nWait, $W_i$ depends on $i$. So for a fixed $j$, we have $\\sum_{i} W_i a_{i,j} \\equiv 0$.\nThis must hold for ALL $j$.\nSo for each column, the weighted sum of elements is 0 mod M.\nThe unknowns are arranged in a grid.\nRow $i$ has value $R_i = \\sum_{j} W_j a_{i,j}$.\nWe know $R_i \\equiv 0$ for $i \\ne k$.\nSo for every row except $k$, the weighted sum is 0.\nFor row $k$, it is not 0.\n\n# Output\nRespond with exactly one lowercase word and nothing else -- no punctuation or\nexplanation:\n- For an incomplete prefix, use `yes` with probability equal to the expected\n normalized rubric credit of a fresh continuation.\n- For a complete response, use `yes` with probability equal to its realized\n normalized rubric credit.\n- Otherwise use `no`.\n"},{"role":"assistant","content":"yes"}],"string":"[\n {\n \"role\": \"user\",\n \"content\": \"# Role\\nYou are forecasting the rubric value of a response trajectory for an olympiad\\nmathematics proof. The problem and response prefix below are fixed. The prefix may\\nbe incomplete, or it may already contain the complete response including its final\\nanswer.\\n\\nIf the prefix is incomplete, imagine starting exactly where it ends and sampling\\none fresh continuation from the same fixed policy and sampling procedure used to\\ncollect the rollout data. The completed response will be graded like an olympiad\\nproof: partial credit is awarded only for correct, rigorous, and clearly justified\\nprogress, and the awarded points are normalized by the maximum available points.\\nTreat your probability of `yes` as the expected normalized rubric credit in\\nthe interval from 0 to 1.\\n\\nIf the prefix is a complete response, judge its realized outcome as written. Treat\\nyour probability of `yes` as that response's realized normalized rubric credit in\\nthe interval from 0 to 1. Do not imagine replacing or continuing an already\\ncomplete response.\\n\\nThis is an eventual rubric-value forecast, not merely a check that the reasoning\\nshown so far looks plausible. A valid prefix can still have low value when the\\nremaining proof is difficult. A flawed prefix can have nonzero value when the\\npolicy can recover. A guessed final answer without a valid proof earns little\\ncredit unless the grading scheme says otherwise.\\n\\nReference attempts, when present, are side information for estimating the policy\\nand problem. They are not part of the fixed prefix and are not visible to the\\nhypothetical continuation.\\n\\n# Problem\\n$10000$ nonzero digits are written in a $100$ -by- $100$ table, one digit per cell. From left to right, each row forms a $100$ -digit integer. From top to bottom, each column forms a $100$ -digit integer. So the rows and columns form $200$ integers (each with $100$ digits), not necessarily distinct. Prove that if at least $199$ of these $200$ numbers are divisible by $2013$ , then all of them are divisible by $2013$ .\\n\\n# Reference attempts\\n## Attempt 1\\nLet $N = 2013$. We are given a $100 \\\\times 100$ table of nonzero digits $d_{i,j} \\\\in \\\\{1, \\\\dots, 9\\\\}$ for $1 \\\\le i, j \\\\le 100$.\\nLet $R_i$ denote the integer formed by the $i$-th row and $C_j$ denote the integer formed by the $j$-th column. Specifically, the digits are concatenated from left-to-right for rows and top-to-bottom for columns. Thus:\\n$$R_i = \\\\sum_{j=1}^{100} d_{i,j} 10^{100-j}$$\\n$$C_j = \\\\sum_{i=1}^{100} d_{i,j} 10^{100-i}$$\\nWe are given that at least 199 of these 200 integers are divisible by $N$. We aim to prove that all 200 integers are divisible by $N$.\\n\\nFirst, observe that $\\\\gcd(10, 2013) = 1$. Thus, any power of 10, denoted $\\\\omega_k = 10^{100-k}$, is coprime to $2013$ and has a multiplicative inverse modulo $2013$.\\nThe problem states that for at least 199 indices pairs $(i,j)$ associated with the 200 integers, the value is $0 \\\\pmod{2013}$. Let $\\\\mathcal{I}$ be the set of indices of the integers that are divisible by $N$. We have $|\\\\mathcal{I}| \\\\ge 199$.\\nThe set of all integers consists of $\\\\{R_1, \\\\dots, R_{100}\\\\} \\\\cup \\\\{C_1, \\\\dots, C_{100}\\\\}$.\\nThe number of non-divisible integers is $200 - |\\\\mathcal{I}| \\\\le 1$.\\nLet $n_r$ be the number of rows not divisible by $N$ and $n_c$ be the number of columns not divisible by $N$. The total number of non-divisible integers is $n_r + n_c \\\\le 1$.\\nThis leaves two possibilities to consider (excluding the trivial case where 0 are non-divisible, in which case the theorem holds):\\n1. Exactly one row fails to be divisible ($n_r = 1$) and all columns are divisible ($n_c = 0$).\\n2. Exactly one column fails to be divisible ($n_c = 1$) and all rows are divisible ($n_r = 0$).\\n\\nBy symmetry (the transpose of the table swaps rows and columns but preserves the structure and divisibility properties), we only need to disprove case 1. If case 1 leads to a contradiction, case 2 leads to the same contradiction by relabeling rows and columns.\\n\\n**Disproving Case 1:**\\nAssume that there exists an index $k \\\\in \\\\{1, \\\\dots, 100\\\\}$ such that $R_k$ is not divisible by $N$, while for all $i \\\\neq k$, $R_i$ is divisible by $N$. Furthermore, assume that all columns $C_j$ are divisible by $N$.\\n\\nThe condition that all columns are divisible means that for every $j \\\\in \\\\{1, \\\\dots, 100\\\\}$:\\n$$C_j = \\\\sum_{i=1}^{100} d_{i,j} 10^{100-i} \\\\equiv 0 \\\\pmod{2013}$$\\nLet us denote the weight associated with the $m$-th position in a number as $\\\\omega_m = 10^{100-m}$. Then the conditions are:\\n1. For all $j$, $\\\\sum_{i=1}^{100} d_{i,j} \\\\omega_i \\\\equiv 0 \\\\pmod{2013}$.\\n2. For all $i \\\\neq k$, $\\\\sum_{j=1}^{100} d_{i,j} \\\\omega_j \\\\equiv 0 \\\\pmod{2013}$.\\n\\nConsider the equation for column $j$ (from condition 1):\\n$$d_{k,j} \\\\omega_k + \\\\sum_{i \\\\neq k} d_{i,j} \\\\omega_i \\\\equiv 0 \\\\pmod{2013}$$\\nSince $\\\\omega_k$ is a unit modulo $2013$ (as $\\\\gcd(10, 2013)=1$), we can isolate $d_{k,j}$:\\n$$d_{k,j} \\\\equiv -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} d_{i,j} \\\\omega_i \\\\pmod{2013}$$\\n\\nNow, consider the expression for $R_k$ (modulo $N$). By condition 2 (applied to $R_k$ itself, though we only know it's non-divisible, we can still manipulate the expression to check its value):\\n$$R_k = \\\\sum_{j=1}^{100} d_{k,j} \\\\omega_j$$\\nSubstitute the expression for $d_{k,j}$ into this sum:\\n$$R_k \\\\equiv \\\\sum_{j=1}^{100} \\\\left( -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} d_{i,j} \\\\omega_i \\\\right) \\\\omega_j \\\\pmod{2013}$$\\nRearranging the summation order:\\n$$R_k \\\\equiv -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} \\\\omega_i \\\\left( \\\\sum_{j=1}^{100} d_{i,j} \\\\omega_j \\\\right) \\\\pmod{2013}$$\\nThe inner sum $\\\\sum_{j=1}^{100} d_{i,j} \\\\omega_j$ is precisely the integer value of row $i$, denoted $R_i$.\\nThus, we have:\\n$$R_k \\\\equiv -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} \\\\omega_i R_i \\\\pmod{2013}$$\\nUnder our assumption for Case 1, $R_i$ is divisible by $N$ for all $i \\\\neq k$. Therefore, $R_i \\\\equiv 0 \\\\pmod{2013}$.\\nSubstituting $R_i \\\\equiv 0$ into the equation:\\n$$R_k \\\\equiv -\\\\omega_k^{-1} \\\\sum_{i \\\\neq k} \\\\omega_i \\\\cdot 0 \\\\equiv 0 \\\\pmod{2013}$$\\nThis implies that $R_k$ is divisible by $2013$. This contradicts the assumption that $R_k$ is not divisible by $2013$.\\n\\nTherefore, it is impossible for exactly one row to fail while all columns succeed. By symmetry, it is also impossible for exactly one column to fail while all rows succeed.\\n\\nConsequently, the only way to satisfy the condition that at least 199 of the 200 integers are divisible by 2013 is if none of them fail. That is, $n_r = 0$ and $n_c = 0$.\\n\\nThus, all 200 integers are divisible by 2013.\\n\\n**Proof that no other answer works (rigorous summary):**\\nWe established that $n_r + n_c \\\\le 1$ is required by the problem statement.\\n1. If $n_r + n_c = 0$, the condition is met trivially.\\n2. If $n_r + n_c = 1$, we proved it leads to a contradiction ($R_k \\\\equiv 0 \\\\pmod{2013}$ if assumed non-zero, or $C_m \\\\equiv 0 \\\\pmod{2013}$ if assumed non-zero).\\nSince the case $n_r + n_c > 1$ yields fewer than 199 divisible numbers (specifically, at most 198), it is excluded by the problem hypothesis.\\nThus, the only valid state is where all 200 integers are divisible.\\n\\nQ.E.D.\\n\\nThe final result is that all 200 numbers must be divisible by 2013.\\n\\nFinal Answer: All 200 of the integers are divisible by 2013.\\n\\n## Attempt 2\\nLet $A = (a_{i,j})$ be the $100 \\\\times 100$ table of nonzero digits, where $1 \\\\le i, j \\\\le 100$.\\nFor each row $i$, let $R_i$ be the integer formed by the digits $a_{i,1}a_{i,2}\\\\dots a_{i,100}$.\\nThe value is given by $R_i = \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j}$.\\nFor each column $j$, let $C_j$ be the integer formed by the digits $a_{1,j}a_{2,j}\\\\dots a_{100,j}$.\\nThe value is given by $C_j = \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\\nLet $M = 2013$. We are given that at least 199 of the integers $R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}$ are divisible by $M$.\\nWe must prove that all 200 integers are divisible by $M$.\\n\\nSuppose, for the sake of contradiction, that at least one integer is not divisible by $M$. Since at most 1 is not divisible (as 199 are divisible), there exists exactly one integer among the 200 that is not divisible by $M$.\\n\\nLet us define weights $w_k = 10^{100-k}$ for $k=1, \\\\dots, 100$.\\nThen $R_i = \\\\sum_{j=1}^{100} a_{i,j} w_j$ and $C_j = \\\\sum_{i=1}^{100} a_{i,j} w_i$.\\nConsider the scalar quantity $S = \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{i,j} w_i w_j$.\\nWe can express $S$ in two ways by grouping the terms:\\n$$S = \\\\sum_{j=1}^{100} w_j \\\\left( \\\\sum_{i=1}^{100} a_{i,j} w_i \\\\right) = \\\\sum_{j=1}^{100} w_j C_j$$\\n$$S = \\\\sum_{i=1}^{100} w_i \\\\left( \\\\sum_{j=1}^{100} a_{i,j} w_j \\\\right) = \\\\sum_{i=1}^{100} w_i R_i$$\\n\\nCase 1: Suppose the non-divisible integer is a column $C_{k_0}$ (with $1 \\\\le k_0 \\\\le 100$).\\nThen $C_j \\\\equiv 0 \\\\pmod M$ for all $j \\\\ne k_0$, and $C_{k_0} \\\\not\\\\equiv 0 \\\\pmod M$.\\nAlso, all rows are divisible, so $R_i \\\\equiv 0 \\\\pmod M$ for all $i$.\\nUsing the expression $S = \\\\sum_{i=1}^{100} w_i R_i$, we substitute $R_i \\\\equiv 0$:\\n$$S \\\\equiv \\\\sum_{i=1}^{100} w_i \\\\cdot 0 \\\\equiv 0 \\\\pmod M.$$\\nUsing the expression $S = \\\\sum_{j=1}^{100} w_j C_j$, we substitute $C_j \\\\equiv 0$ for $j \\\\ne k_0$:\\n$$S \\\\equiv \\\\sum_{j \\\\ne k_0} w_j \\\\cdot 0 + w_{k_0} C_{k_0} \\\\equiv w_{k_0} C_{k_0} \\\\pmod M.$$\\nCombining these, we get $w_{k_0} C_{k_0} \\\\equiv 0 \\\\pmod M$.\\nNote that $w_{k_0} = 10^{100-k_0}$. Since $\\\\gcd(10, M) = \\\\gcd(10, 3 \\\\cdot 11 \\\\cdot 61) = 1$, the weight $w_{k_0}$ is coprime to $M$ and thus is invertible modulo $M$.\\nTherefore, $w_{k_0} C_{k_0} \\\\equiv 0 \\\\implies C_{k_0} \\\\equiv 0 \\\\pmod M$, which contradicts the assumption that $C_{k_0}$ is not divisible by $M$.\\n\\nCase 2: Suppose the non-divisible integer is a row $R_{i_0}$ (with $1 \\\\le i_0 \\\\le 100$).\\nThen $R_i \\\\equiv 0 \\\\pmod M$ for all $i \\\\ne i_0$, and $R_{i_0} \\\\not\\\\equiv 0 \\\\pmod M$.\\nAlso, all columns are divisible, so $C_j \\\\equiv 0 \\\\pmod M$ for all $j$.\\nUsing $S = \\\\sum_{i=1}^{100} w_i R_i$, we substitute $R_i \\\\equiv 0$ for $i \\\\ne i_0$:\\n$$S \\\\equiv w_{i_0} R_{i_0} + \\\\sum_{i \\\\ne i_0} w_i \\\\cdot 0 \\\\equiv w_{i_0} R_{i_0} \\\\pmod M.$$\\nUsing $S = \\\\sum_{j=1}^{100} w_j C_j$, we substitute $C_j \\\\equiv 0$ for all $j$:\\n$$S \\\\equiv \\\\sum_{j=1}^{100} w_j \\\\cdot 0 \\\\equiv 0 \\\\pmod M.$$\\nThis implies $w_{i_0} R_{i_0} \\\\equiv 0 \\\\pmod M$. Again, $w_{i_0}$ is invertible modulo $M$, so $R_{i_0} \\\\equiv 0 \\\\pmod M$, contradicting the assumption that $R_{i_0}$ is not divisible.\\n\\nThus, in either case, the assumption that exactly one integer is not divisible leads to a contradiction.\\nTherefore, it is impossible for exactly 199 integers to be divisible by 2013. Since at least 199 are divisible, all 200 must be divisible by 2013.\\n\\nFinal Result:\\nAll 200 integers formed by the rows and columns are divisible by 2013.\\n\\n## Attempt 3\\nLet $n = 100$ and $M = 2013$.\\nLet $d_{i,j}$ be the digit in the $i$-th row and $j$-th column, where $1 \\\\le i, j \\\\le n$. The problem states that $d_{i,j} \\\\in \\\\{1, 2, \\\\dots, 9\\\\}$.\\nThe integer formed by the $i$-th row is denoted by $R_i$. Based on the convention of writing digits from left to right, $R_i$ is given by:\\n$$ R_i = \\\\sum_{j=1}^{n} d_{i,j} 10^{n-j} $$\\nSimilarly, the integer formed by the $j$-th column is denoted by $C_j$:\\n$$ C_j = \\\\sum_{i=1}^{n} d_{i,j} 10^{n-i} $$\\nWe are given that at least $2n-1$ of these $2n$ integers ($R_1, \\\\dots, R_n, C_1, \\\\dots, C_n$) are divisible by $M$. We must prove that all of them are divisible by $M$.\\n\\nConsider the following algebraic identity involving the row and column sums:\\n$$ S = \\\\sum_{i=1}^{n} R_i 10^{n-i} $$\\nSubstituting the expression for $R_i$:\\n$$ S = \\\\sum_{i=1}^{n} 10^{n-i} \\\\left( \\\\sum_{j=1}^{n} d_{i,j} 10^{n-j} \\\\right) = \\\\sum_{i=1}^{n} \\\\sum_{j=1}^{n} d_{i,j} 10^{2n - i - j} $$\\nSimilarly, consider the sum of the column integers weighted by powers of 10:\\n$$ T = \\\\sum_{j=1}^{n} C_j 10^{n-j} $$\\nSubstituting the expression for $C_j$:\\n$$ T = \\\\sum_{j=1}^{n} 10^{n-j} \\\\left( \\\\sum_{i=1}^{n} d_{i,j} 10^{n-i} \\\\right) = \\\\sum_{j=1}^{n} \\\\sum_{i=1}^{n} d_{i,j} 10^{2n - i - j} $$\\nSince the summation variables $i$ and $j$ run over the same range $[1, n]$, the double sums are identical:\\n$$ \\\\sum_{i=1}^{n} \\\\sum_{j=1}^{n} d_{i,j} 10^{2n - i - j} = \\\\sum_{j=1}^{n} \\\\sum_{i=1}^{n} d_{i,j} 10^{2n - i - j} $$\\nThus, we have the identity:\\n$$ \\\\sum_{i=1}^{n} R_i 10^{n-i} \\\\equiv \\\\sum_{j=1}^{n} C_j 10^{n-j} \\\\pmod M $$\\nThis identity holds as an equality of integers, and consequently as a congruence modulo $M$.\\n\\nLet $S_R = \\\\{ i \\\\in \\\\{1, \\\\dots, n\\\\} \\\\mid R_i \\\\equiv 0 \\\\pmod M \\\\}$ and $S_C = \\\\{ j \\\\in \\\\{1, \\\\dots, n\\\\} \\\\mid C_j \\\\equiv 0 \\\\pmod M \\\\}$.\\nThe condition that at least $199$ of the $200$ numbers are divisible by $M$ implies that $|S_R| + |S_C| \\\\ge 199$.\\nSince $|S_R| \\\\le 100$ and $|S_C| \\\\le 100$, the possible values for $(|S_R|, |S_C|)$ are limited to cases where the number of non-divisible integers is at most 1 (since $200 - 199 = 1$). Specifically, the number of indices not in $S_R \\\\cup S_C$ is $k$ such that $200 - |S_R| - |S_C| \\\\ge 200 - 199 = 1$, so the count of non-divisible numbers is $N_{bad} = 200 - |S_R| - |S_C| \\\\le 1$.\\n\\nWe examine the possible scenarios for the set of \\\"bad\\\" (non-divisible) numbers:\\n1. **Zero bad numbers:** $|S_R| = 100$ and $|S_C| = 100$. In this case, all 200 numbers are divisible by $M$, and we are done.\\n2. **One bad number:** There are two subcases due to the symmetry between rows and columns.\\n * *Subcase 2a:* $R_i \\\\equiv 0 \\\\pmod M$ for all $i$, but $C_n \\\\not\\\\equiv 0 \\\\pmod M$ (so $|S_R|=100, |S_C|=99$).\\n Applying the identity modulo $M$:\\n $$ \\\\sum_{i=1}^{n} R_i 10^{n-i} \\\\equiv \\\\sum_{j=1}^{n} C_j 10^{n-j} \\\\pmod M $$\\n The Left Hand Side (LHS) is:\\n $$ \\\\sum_{i=1}^{n} 0 \\\\cdot 10^{n-i} \\\\equiv 0 \\\\pmod M $$\\n The Right Hand Side (RHS) is:\\n $$ \\\\sum_{j=1}^{n-1} C_j 10^{n-j} + C_n 10^{n-n} $$\\n Since $C_1, \\\\dots, C_{n-1}$ are divisible by $M$, the first part is $0$. The term $10^{n-n} = 10^0 = 1$.\\n So RHS $\\\\equiv C_n \\\\pmod M$.\\n Thus, $0 \\\\equiv C_n \\\\pmod M$, which implies $C_n$ is divisible by $M$. This contradicts the assumption that $C_n$ is bad. Therefore, this case is impossible.\\n * *Subcase 2b:* $C_j \\\\equiv 0 \\\\pmod M$ for all $j$, but $R_n \\\\not\\\\equiv 0 \\\\pmod M$ (so $|S_R|=99, |S_C|=100$).\\n By symmetry with Subcase 2a, we have:\\n $$ \\\\sum_{i=1}^{n} R_i 10^{n-i} \\\\equiv \\\\sum_{j=1}^{n} C_j 10^{n-j} \\\\pmod M $$\\n $$ R_n \\\\cdot 10^0 + 0 \\\\equiv 0 \\\\implies R_n \\\\equiv 0 \\\\pmod M $$\\n Again, this is a contradiction.\\n\\nAny other configuration would involve at least 2 bad numbers (e.g., $R_n$ and $C_n$ both bad, or $R_a$ and $R_b$ both bad), resulting in at most $200-2 = 198$ divisible numbers, which violates the hypothesis.\\n\\nConsequently, the only possible configuration consistent with the hypothesis is the first one: all 200 numbers are divisible by $M$. Thus, the claim is proven.\\n\\n**Answer:**\\nThe integers formed by the rows are $R_i = \\\\sum_{j=1}^{100} d_{i,j} 10^{100-j}$ and the integers formed by the columns are $C_j = \\\\sum_{i=1}^{100} d_{i,j} 10^{100-i}$. Consider the sums:\\n$$ \\\\sum_{i=1}^{100} R_i 10^{100-i} \\\\quad \\\\text{and} \\\\quad \\\\sum_{j=1}^{100} C_j 10^{100-j}. $$\\nExpanding these sums reveals that both are equal to $\\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} d_{i,j} 10^{200-i-j}$. Hence,\\n$$ \\\\sum_{i=1}^{100} R_i 10^{100-i} \\\\equiv \\\\sum_{j=1}^{100} C_j 10^{100-j} \\\\pmod{2013}. $$\\nAssume that at least 199 of the numbers are divisible by 2013. This implies that at most one number is not divisible. If there are no non-divisible numbers, we are done. If there is exactly one non-divisible number, symmetry allows us to assume it is a column integer $C_n$ (the other case is symmetric to considering $R_n$). Then $R_i \\\\equiv 0 \\\\pmod{2013}$ for all $i$ and $C_j \\\\equiv 0 \\\\pmod{2013}$ for all $j < n$. Substituting these into the identity yields:\\n$$ \\\\sum_{i=1}^{100} 0 \\\\cdot 10^{100-i} \\\\equiv \\\\sum_{j=1}^{n-1} 0 \\\\cdot 10^{100-j} + C_n \\\\cdot 10^0 \\\\pmod{2013} $$\\n$$ 0 \\\\equiv C_n \\\\pmod{2013}. $$\\nThis implies $C_n$ is divisible by 2013, contradicting the assumption that it was non-divisible. Therefore, it is impossible to have exactly one non-divisible number. The hypothesis implies that all 200 numbers must be divisible by 2013.\\n\\n## Attempt 4\\nLet the digits in the table be denoted by $a_{ij}$ for $1 \\\\le i, j \\\\le 100$, where $a_{ij} \\\\in \\\\{1, 2, \\\\dots, 9\\\\}$ are the nonzero digits.\\nLet $R_i$ be the integer formed by the $i$-th row, and $C_j$ be the integer formed by the $j$-th column.\\nBased on the positional notation in base 10, we can express these integers as follows:\\n$$ R_i = \\\\sum_{j=1}^{100} a_{ij} 10^{100-j} $$\\n$$ C_j = \\\\sum_{i=1}^{100} a_{ij} 10^{100-i} $$\\nLet us define a sequence of weights $w_k = 10^{100-k}$ for $k = 1, 2, \\\\dots, 100$.\\nThen the expressions become:\\n$$ R_i = \\\\sum_{j=1}^{100} a_{ij} w_j $$\\n$$ C_j = \\\\sum_{i=1}^{100} a_{ij} w_i $$\\nWe consider the quantity $S = \\\\sum_{i=1}^{100} R_i w_i$. Substituting the expression for $R_i$, we get:\\n$$ \\\\begin{aligned} S &= \\\\sum_{i=1}^{100} w_i \\\\left( \\\\sum_{j=1}^{100} a_{ij} w_j \\\\right) \\\\\\\\ &= \\\\sum_{i=1}^{100} \\\\sum_{j=1}^{100} a_{ij} w_i w_j \\\\end{aligned} $$\\nSimilarly, consider the quantity $T = \\\\sum_{j=1}^{100} C_j w_j$. Substituting the expression for $C_j$, we get:\\n$$ \\\\begin{aligned} T &= \\\\sum_{j=1}^{100} w_j \\\\left( \\\\sum_{i=1}^{100} a_{ij} w_i \\\\right) \\\\\\\\ &= \\\\sum_{j=1}^{100} \\\\sum_{i=1}^{100} a_{ij} w_j w_i \\\\end{aligned} $$\\nSince finite sums commute and multiplication of integers is commutative ($w_i w_j = w_j w_i$), we observe that $S = T$. That is:\\n$$ \\\\sum_{i=1}^{100} R_i w_i = \\\\sum_{j=1}^{100} C_j w_j $$\\nThis equality holds over the integers. Therefore, it also holds modulo any integer $m$. In particular, let $m = 2013$.\\n$$ \\\\sum_{i=1}^{100} R_i w_i \\\\equiv \\\\sum_{j=1}^{100} C_j w_j \\\\pmod{2013} $$\\nWe are given that at least 199 of the 200 integers $\\\\{R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}\\\\}$ are divisible by 2013. This means there is at most 1 integer among them that is not divisible by 2013.\\nWe proceed by contradiction. Suppose there exists exactly one integer $X$ in the set which is not divisible by 2013. There are two cases:\\nCase 1: The non-divisible integer is a row, say $R_k$ for some $k \\\\in \\\\{1, \\\\dots, 100\\\\}$.\\nThen $R_i \\\\equiv 0 \\\\pmod{2013}$ for all $i \\\\neq k$, and $C_j \\\\equiv 0 \\\\pmod{2013}$ for all $j \\\\in \\\\{1, \\\\dots, 100\\\\}$.\\nSubstituting these into our congruence relation:\\n$$ \\\\sum_{i=1}^{100} R_i w_i \\\\equiv R_k w_k + \\\\sum_{i \\\\neq k} 0 \\\\cdot w_i \\\\equiv R_k w_k \\\\pmod{2013} $$\\n$$ \\\\sum_{j=1}^{100} C_j w_j \\\\equiv \\\\sum_{j=1}^{100} 0 \\\\cdot w_j \\\\equiv 0 \\\\pmod{2013} $$\\nEquating these results gives:\\n$$ R_k w_k \\\\equiv 0 \\\\pmod{2013} $$\\nRecall that $w_k = 10^{100-k}$. The modulus is $2013$. We factorize 2013 as $2013 = 3 \\\\times 11 \\\\times 61$.\\nSince $10$ is not divisible by 3, 11, or 61 (obvious inspection or calculation), $\\\\gcd(10, 2013) = 1$. Consequently, $\\\\gcd(w_k, 2013) = \\\\gcd(10^{100-k}, 2013) = 1$.\\nThis implies that $w_k$ is invertible modulo 2013. Multiplying both sides of $R_k w_k \\\\equiv 0 \\\\pmod{2013}$ by the modular inverse of $w_k$, we obtain:\\n$$ R_k \\\\equiv 0 \\\\pmod{2013} $$\\nThis contradicts the assumption that $R_k$ is not divisible by 2013.\\n\\nCase 2: The non-divisible integer is a column, say $C_k$ for some $k \\\\in \\\\{1, \\\\dots, 100\\\\}$.\\nThen $R_i \\\\equiv 0 \\\\pmod{2013}$ for all $i$, and $C_j \\\\equiv 0 \\\\pmod{2013}$ for all $j \\\\neq k$.\\nSubstituting these into the congruence relation:\\n$$ \\\\sum_{i=1}^{100} R_i w_i \\\\equiv 0 \\\\pmod{2013} $$\\n$$ \\\\sum_{j=1}^{100} C_j w_j \\\\equiv C_k w_k + \\\\sum_{j \\\\neq k} 0 \\\\cdot w_j \\\\equiv C_k w_k \\\\pmod{2013} $$\\nEquating these gives:\\n$$ C_k w_k \\\\equiv 0 \\\\pmod{2013} $$\\nAs before, since $\\\\gcd(w_k, 2013) = 1$, it follows that $C_k \\\\equiv 0 \\\\pmod{2013}$.\\nThis contradicts the assumption that $C_k$ is not divisible by 2013.\\n\\nSince both cases lead to a contradiction, our initial assumption that there is exactly one non-divisible integer must be false.\\nThus, there are either 0 or fewer than 1 non-divisible integers. Given that \\\"at least 199\\\" are divisible, it is impossible to have fewer than 199 divisible. If exactly 1 failed, we have a contradiction. Thus, 0 fail.\\nTherefore, all 200 integers $R_1, \\\\dots, R_{100}$ and $C_1, \\\\dots, C_{100}$ are divisible by 2013.\\n\\n# Fixed response prefix\\nLet the $100 \\\\times 100$ table be denoted by $A = (a_{i,j})$ where $1 \\\\le i, j \\\\le 100$.\\nThe digits are nonzero, so $a_{i,j} \\\\in \\\\{1, 2, \\\\dots, 9\\\\}$. Wait, the problem says \\\"nonzero digits\\\".\\nThis means $a_{i,j} \\\\in \\\\{1, \\\\dots, 9\\\\}$? Or could it be that they are just integers $1, \\\\dots, 9$? \\\"Digits\\\" usually implies base 10 digits. Nonzero means $\\\\{1, \\\\dots, 9\\\\}$.\\n\\nLet $R_i$ be the integer formed by the $i$-th row for $1 \\\\le i \\\\le 100$.\\nLet $C_j$ be the integer formed by the $j$-th column for $1 \\\\le j \\\\le 100$.\\nWe have a set of $N = 200$ integers $\\\\{R_1, \\\\dots, R_{100}, C_1, \\\\dots, C_{100}\\\\}$.\\nLet $S$ be this set (as a multiset of values). The size is 200.\\nWe are given that at least 199 of these numbers are divisible by $2013$.\\nWe need to prove that all of them are divisible by $2013$.\\nThis is equivalent to: If $| \\\\{ x \\\\in S : x \\\\equiv 0 \\\\pmod{2013} \\\\} | \\\\ge 199$, then for all $x \\\\in S$, $x \\\\equiv 0 \\\\pmod{2013}$.\\n\\nLet's check the modulus. $2013 = 3 \\\\times 11 \\\\times 61$.\\nWait, $2+0+1+3 = 6$, so divisible by 3. $2013 / 3 = 671$.\\n$671 = 11 \\\\times 61$. So $2013 = 3 \\\\times 11 \\\\times 61$.\\nThis factorization might not be crucial immediately, but good to know.\\n\\nLet's denote $r(i)$ as the remainder of $R_i$ modulo 2013, and $c(j)$ as the remainder of $C_j$ modulo 2013.\\nThere are $m=100$ rows and $n=100$ columns.\\nLet $r_i \\\\equiv R_i \\\\pmod m$ and $c_j \\\\equiv C_j \\\\pmod m$. Actually the modulus is $M=2013$.\\nSo $R_i = \\\\sum_{j=1}^{100} a_{i,j} 10^{100-j}$ and $C_j = \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\\nThe condition is that $r_i = 0$ or $c_j = 0$ for almost all indices.\\nSuppose exactly one number is not divisible by 2013. Without loss of generality, assume $R_k$ is not divisible, and all other $R_i$ ($i \\\\ne k$) and all $C_j$ ($1 \\\\le j \\\\le 100$) are divisible.\\nOr maybe two are not divisible, etc. But we want to show that the assumption of $\\\\ge 199$ implies all are divisible. So let's try to construct a contradiction if there is one non-divisible number.\\nOr rather, let's analyze the sum of remainders or relations between $R_i$'s and $C_j$'s.\\nSpecifically, $R_i$ depends on the row entries, $C_j$ depends on column entries. They intersect at $(i,j)$.\\n\\nLet $x_{i,j} = a_{i,j}$. These are variables in $\\\\{1, \\\\dots, 9\\\\}$. This constraint might be used to prevent trivial counterexamples (e.g., setting some digits to make things cancel out mod M easily? No, the range is restricted).\\nHowever, $100$ digits forming an integer means $R_i = \\\\sum_{j=1}^{100} x_{i,j} 10^{100-j}$.\\nNotice that $10^{100} \\\\equiv 1 \\\\pmod 9$. Modulo 3, $10 \\\\equiv 1$.\\nModulo 11, $10 \\\\equiv -1$.\\nModulo 61...\\n\\nLet's look at the structure of the linear map from the space of matrices to the values modulo $M$.\\nLet $\\\\phi(M)$ be the map $(a_{i,j}) \\\\mapsto (\\\\text{values } R_i, C_j \\\\bmod M)$.\\nThis looks like we are trying to solve a system of linear congruences over the matrix variables.\\nBut the number of equations is 200, number of variables is 10000.\\nActually, the divisibility condition is about the specific values.\\nIs there a simpler way?\\nConsider the sum of the remainders. Or sum of digits.\\nIf all numbers are divisible by 3, sum of digits of each must be divisible by 3.\\nFor rows, $\\\\sum_{j=1}^{100} a_{i,j}$ is sum of digits of $R_i$.\\nSince $R_i$ is formed by digits $a_{i,j}$, $R_i \\\\equiv \\\\sum a_{i,j} \\\\pmod 9$.\\nWait, $M=2013$ is not 9. It is divisible by 3. So $R_i \\\\equiv 0 \\\\pmod 3 \\\\iff \\\\sum a_{i,j} \\\\equiv 0 \\\\pmod 3$.\\nSimilarly for columns.\\nIf at least 199 integers are divisible by 2013, they are certainly divisible by 3.\\nThis imposes constraints on the sums of digits in rows and columns.\\nLet $s_{i} = \\\\sum_{j=1}^{100} a_{i,j}$ and $t_{j} = \\\\sum_{i=1}^{100} a_{i,j}$.\\n$s_i$ is the sum of digits in row $i$. $t_j$ is sum of digits in column $j$.\\n$\\\\sum s_i = \\\\sum t_j = \\\\sum a_{i,j}$.\\nIf $R_i \\\\equiv 0 \\\\pmod 3$, then $s_i \\\\equiv 0 \\\\pmod 3$.\\nIf $C_j \\\\equiv 0 \\\\pmod 3$, then $t_j \\\\equiv 0 \\\\pmod 3$.\\nSuppose exactly one row $R_k$ is NOT divisible by 2013. Then all $R_i$ ($i \\\\ne k$) are divisible by 2013 $\\\\implies$ divisible by 3.\\nAll $C_j$ are divisible by 2013 $\\\\implies$ divisible by 3.\\nThen $s_i \\\\equiv 0 \\\\pmod 3$ for $i \\\\ne k$, and $t_j \\\\equiv 0 \\\\pmod 3$ for all $j$.\\nWhat about $s_k$?\\n$\\\\sum_{j=1}^{100} a_{k,j} = s_k$. Also $s_k = \\\\sum_{j} a_{k,j}$.\\nFrom $t_j \\\\equiv 0 \\\\pmod 3$, we have $\\\\sum_{i} a_{i,j} \\\\equiv 0 \\\\pmod 3$.\\nSumming over $j$: $\\\\sum_{j} \\\\sum_{i} a_{i,j} = \\\\sum s_i \\\\equiv 0 \\\\pmod 3$.\\nWe have $s_k + \\\\sum_{i \\\\ne k} s_i \\\\equiv 0 \\\\pmod 3$.\\nSince $s_i \\\\equiv 0 \\\\pmod 3$ for $i \\\\ne k$, this gives $s_k \\\\equiv 0 \\\\pmod 3$.\\nThus $s_k \\\\equiv 0 \\\\pmod 3$.\\nSo $R_k \\\\equiv \\\\sum a_{k,j} \\\\equiv 0 \\\\pmod 3$.\\nThis doesn't help with divisibility by 2013 directly because $2013$ has factors 11 and 61.\\nBut $R_k \\\\equiv 0 \\\\pmod 3$ is necessary. Is it possible that $R_k$ is not divisible by 2013 but is divisible by 3? Yes.\\nHowever, the condition is stronger. 199 out of 200 are divisible by 2013.\\nLet's see. Suppose $R_k$ is the only one not divisible.\\nThen $R_k \\\\not\\\\equiv 0 \\\\pmod{2013}$.\\nWe established that $R_k \\\\equiv 0 \\\\pmod 3$.\\nSimilarly, consider modulo 11. $2013 = 3 \\\\cdot 11 \\\\cdot 61$.\\nIf 199 numbers are divisible by $M$, they are divisible by 11.\\nIf only one number $x$ is not divisible by $M$, it must be divisible by 11 or not?\\nWait. If $x$ is not divisible by $M$, it means $x \\\\not\\\\equiv 0 \\\\pmod{M}$.\\nThis does NOT imply $x \\\\not\\\\equiv 0 \\\\pmod p$ for any factor $p$.\\nHowever, the hypothesis says \\\"at least 199 ... are divisible by 2013\\\".\\nLet $A$ be the set of numbers. Let $Z$ be the subset of numbers divisible by 2013.\\n$|Z| \\\\ge 199$. So the complement $Z^c$ has size $\\\\le 1$.\\nCase 1: All numbers are divisible. Done.\\nCase 2: There is exactly one number not divisible.\\nWithout loss of generality, suppose it is $R_k$. Then all $R_i$ ($i \\\\ne k$) and all $C_j$ are divisible by 2013.\\nIn particular, they are divisible by 11.\\n$R_i \\\\equiv \\\\sum_{j=1}^{100} a_{i,j} (-1)^{100-j} \\\\equiv \\\\sum_{j=1}^{100} a_{i,j} (-1)^{j} \\\\pmod{11}$ since $(-1)^{100}=1$.\\nWait, digit $d$ at position $j$ (from left, $j=1$) has weight $10^{100-j}$.\\n$j=1 \\\\implies 10^{99} \\\\equiv (-1)^{99} \\\\equiv -1 \\\\pmod{11}$.\\n$j=100 \\\\implies 10^0 = 1 \\\\equiv 1 \\\\pmod{11}$.\\nSo $R_i \\\\equiv \\\\sum_{j=1}^{100} a_{i,j} (-1)^{100-j} \\\\pmod{11}$.\\nSince $100-j$ is even for even $j$, odd for odd $j$.\\nSo $R_i \\\\equiv \\\\sum_{j=1}^{100} (-1)^{j+1} a_{i,j} \\\\pmod{11}$ ? Let's check signs.\\n$100-1=99$ odd $\\\\to -1$. $100-2=98$ even $\\\\to 1$.\\nSo coeff of $a_{i,j}$ is $(-1)^{100-j}$.\\nLet's denote $w_j = (-1)^{100-j}$. Note $w_j = (-1)^j w_1$. $w_1 = -1$. So $w_j = -(-1)^j = (-1)^{j+1}$.\\nActually let's just write $\\\\sum_{j} (-1)^{j} a_{i,j}$ with adjusted sign.\\nLet's use alternating sum of digits.\\nLet $A_i$ be the alternating sum of row $i$. $A_i = \\\\sum_{j=1}^{100} (-1)^{j-1} a_{i,j}$. Or something similar.\\nDivisibility by 11 depends on this.\\nIf all $C_j$ are divisible by 11, then the alternating sum of each column is divisible by 11.\\nIf all $R_i$ for $i \\\\ne k$ are divisible by 11, then their alternating sums are divisible by 11.\\nDoes this force $R_k$ to be divisible by 11?\\nLet's define $L(R) = \\\\sum_{j=1}^{100} (-1)^{j} a_{1,j} + \\\\dots$. No, row index $i$.\\n$L_i = \\\\sum_{j=1}^{100} (-1)^j a_{i,j}$. (Using some fixed convention, say $100-1$ is odd power of 10).\\nLet's fix the weights: $P_j = 10^{100-j}$.\\n$R_i = \\\\sum_{j=1}^{100} P_j a_{i,j}$.\\n$C_j = \\\\sum_{i=1}^{100} P_i a_{i,j}$ (Wait, column digits are ordered top to bottom, so position $i$ has weight $10^{100-i}$).\\nYes, $C_j = \\\\sum_{i=1}^{100} a_{i,j} 10^{100-i}$.\\nNote the weights are the same function of index. Let $W_k = 10^{100-k}$.\\n$R_i = \\\\sum_{j=1}^{100} W_j a_{i,j}$.\\n$C_j = \\\\sum_{i=1}^{100} W_i a_{i,j}$.\\n\\nWe have 100 conditions for $C_j$ being 0 mod 2013, and 100 conditions for $R_i$ being 0 mod 2013.\\nActually, we have 199 conditions satisfied.\\nAssume $R_k$ is the exceptional one. So $R_k \\\\not\\\\equiv 0$.\\nAll $C_j \\\\equiv 0 \\\\pmod{2013}$.\\nAlso $R_i \\\\equiv 0 \\\\pmod{2013}$ for all $i \\\\ne k$.\\nConsider the determinant or something? No.\\nLook at the intersection of sets of constraints.\\nLet's work in the ring $\\\\mathbb{Z}_{2013}$.\\nThe $C_j$'s give information about the weighted column sums.\\nThe $R_i$'s give info about weighted row sums.\\nIs it possible that we can deduce something about the structure of the matrix?\\nThe digits $a_{i,j}$ are just numbers.\\nMaybe we can sum $R_i$ or something.\\nOr sum $C_j$.\\nBut that sums everything up twice.\\nConsider $\\\\sum_{i} R_i = \\\\sum_{i} \\\\sum_{j} W_j a_{i,j} = \\\\sum_{j} W_j (\\\\sum_{i} a_{i,j}) = \\\\sum_{j} W_j T_j$ where $T_j$ is sum of col $j$.\\nThis doesn't seem to link $R$ and $C$ strongly.\\nHowever, we know $R_i$ and $C_j$ individually.\\n\\nLet's look at the modular arithmetic properties.\\nThe modulus is $M = 2013 = 3 \\\\cdot 11 \\\\cdot 61$.\\nIt turns out that if a set of numbers are multiples of $M$, then certain algebraic relations must hold.\\nPerhaps we can show that $R_k \\\\equiv \\\\text{something derived from } C_j \\\\text{ and } R_i$?\\nWait, the $C_j$ are fully known (they are 0 mod M).\\nSo $C_j \\\\equiv 0$ for all $j$.\\nThis means $\\\\sum_{i=1}^{100} a_{i,j} W_i \\\\equiv 0 \\\\pmod M$.\\nHere $a_{i,j}$ are unknowns.\\nThe equation $C_j \\\\equiv 0$ holds for $j=1,\\\\dots,100$.\\nThis is a system of 100 linear equations in $a_{i,j}$ (with coefficients $W_i$).\\nWait, $W_i$ depends on $i$. So for a fixed $j$, we have $\\\\sum_{i} W_i a_{i,j} \\\\equiv 0$.\\nThis must hold for ALL $j$.\\nSo for each column, the weighted sum of elements is 0 mod M.\\nThe unknowns are arranged in a grid.\\nRow $i$ has value $R_i = \\\\sum_{j} W_j a_{i,j}$.\\nWe know $R_i \\\\equiv 0$ for $i \\\\ne k$.\\nSo for every row except $k$, the weighted sum is 0.\\nFor row $k$, it is not 0.\\n\\n# Output\\nRespond with exactly one lowercase word and nothing else -- no punctuation or\\nexplanation:\\n- For an incomplete prefix, use `yes` with probability equal to the expected\\n normalized rubric credit of a fresh continuation.\\n- For a complete response, use `yes` with probability equal to its realized\\n normalized rubric credit.\\n- Otherwise use `no`.\\n\"\n },\n {\n \"role\": \"assistant\",\n \"content\": \"yes\"\n }\n]","isConversational":true},"enable_thinking":{"kind":"bool","value":false,"string":"false"},"id":{"kind":"string","value":"8cf1cb3370041166"},"source":{"kind":"string","value":"fineproofs"},"sample_index":{"kind":"number","value":7,"string":"7"},"step_idx":{"kind":"number","value":8,"string":"8"},"num_steps":{"kind":"number","value":80,"string":"80"},"label":{"kind":"string","value":"yes"},"correct":{"kind":"bool","value":true,"string":"true"},"loss_weight":{"kind":"number","value":1,"string":"1"},"reward":{"kind":"number","value":0.9464285714285714,"string":"0.946429"}}}],"truncated":false,"partial":false},"paginationData":{"pageIndex":0,"numItemsPerPage":100,"numTotalItems":53457,"offset":0,"length":100}},"jwt":"eyJhbGciOiJFZERTQSIsImtpZCI6IjVHZDBvd0g5MTM2eDZjc1FvbE1zcktNWUZoRVFoUm5PVVVybEpjOUhaUEEifQ.eyJyZWFkIjp0cnVlLCJwZXJtaXNzaW9ucyI6eyJyZXBvLmNvbnRlbnQucmVhZCI6dHJ1ZX0sInJlcG9JZHMiOlsiNmE2ZjVhOTljMDFlNDEyYjQ3OTIxMWI1Il0sImlhdCI6MTc5MTE0NjkzMCwianRpIjoiNGE0OTgzOGQtODhlZS00Y2QzLTk2ZjctZTg5YjkyYmExZWY5Iiwic3ViIjoiL2RhdGFzZXRzL2FzaW5naDE1L2ZpbmVwcm9vZnMtcHJtLWNvbnRleHQtdjItc29sdXRpb24iLCJleHAiOjE3OTExNTA1MzAsImlzcyI6Imh0dHBzOi8vaHVnZ2luZ2ZhY2UuY28ifQ.uK7Zpeir5GnEv_Xp-6JPfXFpc6AveYxzonxn0-lEQtqoCfU7Lq7RnZrgojm1fVNgbFP4ptfUnOT7OVvjkXZ5BA","displayUrls":true,"splitSizeSummaries":[{"config":"default","split":"train","numRows":53457,"numBytesParquet":2087341563},{"config":"default","split":"validation","numRows":2602,"numBytesParquet":105707434}]},"discussionsStats":{"closed":0,"open":0,"total":0},"fullWidth":true,"hasGatedAccess":true,"hasFullAccess":true,"isEmbedded":false,"savedQueries":{"community":[],"user":[]}}">
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"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 5 | 4 | 14 | yes | false | 1 | 0.433036 |
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"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 5 | 12 | 14 | yes | false | 1 | 0.03125 |
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"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 5 | 14 | 14 | yes | false | 1 | 0 |
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{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 4 | 33 | yes | false | 1 | 0.1875 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 8 | 33 | yes | false | 1 | 0 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 12 | 33 | yes | false | 1 | 0 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 16 | 33 | yes | false | 1 | 0 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 20 | 33 | yes | false | 1 | 0.03125 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 24 | 33 | yes | false | 1 | 0 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 28 | 33 | yes | false | 1 | 0 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 32 | 33 | yes | false | 1 | 0.053571 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | d3b87a033b6e837b | fineproofs | 6 | 33 | 33 | yes | false | 1 | 0.035714 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 4 | 34 | yes | true | 1 | 0.732143 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 8 | 34 | yes | true | 1 | 0.723214 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 12 | 34 | yes | true | 1 | 0.821429 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 16 | 34 | yes | true | 1 | 0.785714 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 20 | 34 | yes | true | 1 | 0.821429 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 24 | 34 | yes | true | 1 | 0.772321 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 28 | 34 | yes | true | 1 | 0.794643 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 32 | 34 | yes | true | 1 | 0.803571 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 2 | 34 | 34 | yes | true | 1 | 0.8125 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 4 | 40 | yes | true | 1 | 0.763393 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 8 | 40 | yes | true | 1 | 0.745536 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 12 | 40 | yes | true | 1 | 0.633929 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 16 | 40 | yes | true | 1 | 0.540179 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 20 | 40 | yes | true | 1 | 0.566964 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 24 | 40 | yes | true | 1 | 0.522321 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 28 | 40 | yes | true | 1 | 0.647321 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 32 | 40 | yes | true | 1 | 0.808036 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 36 | 40 | yes | true | 1 | 0.861607 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 5d2121411de18a19 | fineproofs | 5 | 40 | 40 | yes | true | 1 | 0.799107 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 0 | 4 | 29 | yes | true | 1 | 0.8125 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 0 | 8 | 29 | yes | true | 1 | 0.825893 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 0 | 12 | 29 | yes | true | 1 | 0.84375 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 0 | 16 | 29 | yes | true | 1 | 0.821429 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 0 | 20 | 29 | yes | true | 1 | 0.803571 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 0 | 24 | 29 | yes | true | 1 | 0.825893 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 0 | 28 | 29 | yes | true | 1 | 0.821429 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 0 | 29 | 29 | yes | true | 1 | 0.839286 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 4 | 34 | yes | true | 1 | 0.84375 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 8 | 34 | yes | true | 1 | 0.821429 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 12 | 34 | yes | true | 1 | 0.84375 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 16 | 34 | yes | true | 1 | 0.84375 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 20 | 34 | yes | true | 1 | 0.830357 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 24 | 34 | yes | true | 1 | 0.834821 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 28 | 34 | yes | true | 1 | 0.852679 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 32 | 34 | yes | true | 1 | 0.839286 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 4ec443ed98171332 | fineproofs | 6 | 34 | 34 | yes | true | 1 | 0.821429 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | cbcd6667f33ce082 | fineproofs | 0 | 4 | 6 | yes | true | 1 | 0.946429 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | cbcd6667f33ce082 | fineproofs | 0 | 6 | 6 | yes | true | 1 | 0.924107 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | cbcd6667f33ce082 | fineproofs | 2 | 4 | 22 | yes | true | 1 | 0.888393 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | cbcd6667f33ce082 | fineproofs | 2 | 8 | 22 | yes | true | 1 | 0.861607 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | cbcd6667f33ce082 | fineproofs | 2 | 12 | 22 | yes | true | 1 | 0.901786 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | cbcd6667f33ce082 | fineproofs | 2 | 16 | 22 | yes | true | 1 | 0.919643 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | cbcd6667f33ce082 | fineproofs | 2 | 20 | 22 | yes | true | 1 | 0.924107 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | cbcd6667f33ce082 | fineproofs | 2 | 22 | 22 | yes | true | 1 | 0.892857 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 0 | 4 | 29 | yes | true | 1 | 0.799107 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 0 | 8 | 29 | yes | false | 1 | 0.392857 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 0 | 12 | 29 | yes | true | 1 | 0.616071 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 0 | 16 | 29 | yes | false | 1 | 0.419643 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 0 | 20 | 29 | yes | false | 1 | 0.388393 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 0 | 24 | 29 | yes | false | 1 | 0.410714 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 0 | 28 | 29 | yes | true | 1 | 0.566964 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 0 | 29 | 29 | yes | true | 1 | 0.669643 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 3 | 4 | 28 | yes | true | 1 | 0.544643 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 3 | 8 | 28 | yes | true | 1 | 0.517857 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 3 | 12 | 28 | yes | true | 1 | 0.669643 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 3 | 16 | 28 | yes | false | 1 | 0.473214 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 3 | 20 | 28 | yes | false | 1 | 0.397321 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 3 | 24 | 28 | yes | false | 1 | 0.401786 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 593f05e3f19ff10d | fineproofs | 3 | 28 | 28 | yes | false | 1 | 0.40625 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 0 | 4 | 10 | yes | true | 1 | 0.839286 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 0 | 8 | 10 | yes | true | 1 | 0.986607 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 0 | 10 | 10 | yes | true | 1 | 0.982143 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 4 | 42 | yes | true | 1 | 0.834821 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 8 | 42 | yes | true | 1 | 0.776786 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 12 | 42 | yes | true | 1 | 0.941964 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 16 | 42 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 20 | 42 | yes | true | 1 | 0.995536 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 24 | 42 | yes | true | 1 | 0.991071 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 28 | 42 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 32 | 42 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 36 | 42 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 40 | 42 | yes | true | 1 | 0.991071 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 58690d9e85fa0817 | fineproofs | 2 | 42 | 42 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 4 | 45 | yes | true | 1 | 0.973214 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 8 | 45 | yes | true | 1 | 0.964286 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 12 | 45 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 16 | 45 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 20 | 45 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 24 | 45 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 28 | 45 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 32 | 45 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 36 | 45 | yes | true | 1 | 0.995536 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 40 | 45 | yes | true | 1 | 0.995536 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 44 | 45 | yes | true | 1 | 1 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 1 | 45 | 45 | yes | true | 1 | 0.995536 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 7 | 4 | 80 | yes | true | 1 | 0.915179 |
[
{
"role": "user",
"content": "# Role\nYou are forecasting the rubric value of a response trajectory for an olympiad\nmathematics proof. The problem and response prefix below are fixed. The prefix may\nbe incomplete, or it may already contain the complete response including its final\nanswer.\n\nIf the prefi... | false | 8cf1cb3370041166 | fineproofs | 7 | 8 | 80 | yes | true | 1 | 0.946429 |
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