| const RUNS=[{"q": "What is the sum of the digits of 2^1000?", "gold": "1366", "none": ["**Final answer:** `(no boxed answer)` \n**Generated tokens:** 1,024 \u00b7 **Time:** 61.5 s \u00b7 **Model calls:** 1 \u00b7 **Tool calls:** 0 \n\u26a0\ufe0f hit the token cap", "To find the sum of the digits of \\(2^{1000}\\), we need to determine the decimal representation of \\(2^{1000}\\) and then add its digits. However, calculating \\(2^{1000}\\) directly is impractical by hand due to its large size (it has 302 digits). Instead, we can use properties of modular arithmetic or known results about the digit sum.\n\nThe sum of the digits of a number is congruent to the number modulo 9. This is because the digital root (and thus the sum of digits) is related to \\(n \\mod 9\\). Specifically:\n\\[\n\\text{Sum of digits} \\equiv n \\pmod{9}\n\\]\nBut this only gives us the value modulo 9, not the actual sum. For large exponents like \\(2^{1000}\\), we can compute \\(2^{1000} \\mod 9\\) to find a clue about the digit sum.\n\nFirst, note that:\n\\[\n2^1 = 2 \\equiv 2 \\pmod{9}\n\\]\n\\[\n2^2 = 4 \\equiv 4 \\pmod{9}\n\\]\n\\[\n2^3 = 8 \\equiv 8 \\pmod{9}\n\\]\n\\[\n2^4 = 16 \\equiv 7 \\pmod{9} \\quad (\\text{since } 16 - 9 = 7)\n\\]\n\\[\n2^5 = 32 \\equiv 5 \\pmod{9} \\quad (32 - 27 = 5)\n\\]\n\\[\n2^6 = 64 \\equiv 1 \\pmod{9} \\quad (64 - 63 = 1)\n\\]\nThe powers of 2 modulo 9 cycle every 6: \\(2, 4, 8, 7, 5, 1\\). Since \\(1000 \\div 6 = 166\\) remainder \\(4\\), we have:\n\\[\n2^{1000} \\equiv 2^4 \\equiv 16 \\equiv 7 \\pmod{9}\n\\]\nSo, the sum of the digits of \\(2^{1000}\\) is congruent to 7 modulo 9. That is, if \\(S\\) is the sum of the digits, then:\n\\[\nS \\equiv 7 \\pmod{9}\n\\]\n\nNow, we need the actual value or a way to compute it. There are known results for the digit sums of powers of 2. Alternatively, we can use computational tools or properties. Since this is a text-based reasoning, I recall that \\(2^{1000}\\) has been computed and its digits summed before. However, let's verify using another approach.\n\nWe can also compute the number of digits: \n\\[\n\\log_{10}(2^{1000}) = 1000 \\log_{10} 2 \\approx 1000 \\times 0.30102999566 = 301.02999566\n\\]\nSo \\(2^{1000}\\) has \\(\\lfloor 301.02999566 \\rfloor + 1 = 302\\) digits.\n\nTo find the digit sum, one might need to look up or compute it. Since this is a known problem, I remember that the sum of the digits of \\(2^{1000}\\) is 1367? But let's not guess; we can reason differently.\n\nAlternatively, there is a method using programming or known sequences. For the purpose of this problem, since it's likely expecting an answer from standard knowledge or calculation, and given the modulo 9 result is 7, but that doesn't give the sum directly.\n\nWait, perhaps I can find \\(2^{1000}\\) mod 99 to get more info? But digit sum isn't directly related to mod 99. Another idea: use the fact that for large numbers, the digit sum can be found via software. Since this is a text-based interaction and I need to provide an answer, I'll rely on known results or compute step by step.\n\nUpon recalling or using computational knowledge (as in typical contest problems), the sum of digits of \\(2^{1000}\\) is 1367? Let me check consistency: if S \u2261 7 mod 9, then 1367 \u00f7 9 = 151 * 9 = 1359, remainder 8. That's not 7. So that's wrong.\n\nLet me compute \\(2^{10} = 1024\\), sum of digits = 1+0+2+4=7.\n\\(2^{20}\\): \\(2^{10}^2 = 1024^2 = 1048576\\), sum: 1+0+4+8+5+7+6=31? Wait, better to use known values.\n\nActually, I can compute the digit sum for smaller powers and see a pattern, but it's not straightforward. Given time constraints, I'll assume that the correct answer is known or derived from standard sources. \n\nAlternatively, we can use the fact that \\(2^{1000} = (2^{10})^{100}\\). But still large.\n\nI found online resources or", "_no tool calls_"], "python": ["**Final answer:** `1366` \n**Generated tokens:** 153 \u00b7 **Time:** 12.3 s \u00b7 **Model calls:** 2 \u00b7 **Tool calls:** 1", "The sum of the digits of \\(2^{1000}\\) is \\(\\boxed{1366}\\).", "**run_python #1**\n```python\nn = 2**1000\ns = str(n)\ndigit_sum = sum(int(ch) for ch in s)\nprint(digit_sum)\n```\n```text\n1366\n```"]}, {"q": "What is the remainder when 7^2026 is divided by 1000?", "gold": "649", "none": ["**Final answer:** `(no boxed answer)` \n**Generated tokens:** 1,024 \u00b7 **Time:** 61.4 s \u00b7 **Model calls:** 1 \u00b7 **Tool calls:** 0 \n\u26a0\ufe0f hit the token cap", "To find the remainder when \\(7^{2026}\\) is divided by 1000, we need to compute \\(7^{2026} \\mod 1000\\). Since \\(1000 = 8 \\times 125\\) and 8 and 125 are coprime, we can use the Chinese Remainder Theorem (CRT) by finding the remainders modulo 8 and modulo 125 separately.\n\n**Step 1: Compute \\(7^{2026} \\mod 8\\)**\n\nNote that:\n- \\(7 \\equiv -1 \\pmod{8}\\)\n- So, \\(7^{2026} \\equiv (-1)^{2026} \\pmod{8}\\)\n\nSince 2026 is even, \\((-1)^{2026} = 1\\). Thus,\n\\[\n7^{2026} \\equiv 1 \\pmod{8}\n\\]\n\n**Step 2: Compute \\(7^{2026} \\mod 125\\)**\n\nWe need to find the exponent modulo \\(\\phi(125)\\), where \\(\\phi\\) is Euler's totient function. Since 125 is a power of an odd prime, \\(\\phi(125) = 125 \\times (1 - \\frac{1}{5}) = 100\\).\n\nThus,\n\\[\n7^{2026} \\equiv 7^{2026 \\mod 100} \\pmod{125}\n\\]\nCalculate \\(2026 \\div 100\\):\n- \\(100 \\times 20 = 2000\\), so \\(2026 - 2000 = 26\\). Thus, \\(2026 \\equiv 26 \\pmod{100}\\).\n\nSo,\n\\[\n7^{2026} \\equiv 7^{26} \\pmod{125}\n\\]\n\nNow compute \\(7^{26} \\mod 125\\) using successive squaring:\n\n- \\(7^1 = 7\\)\n- \\(7^2 = 49\\)\n- \\(7^4 = (7^2)^2 = 49^2 = 2401 \\equiv 2401 - 19 \\times 125 = 2401 - 2375 = 26 \\pmod{125}\\)\n- \\(7^8 = (7^4)^2 = 26^2 = 676 \\equiv 676 - 5 \\times 125 = 676 - 625 = 51 \\pmod{125}\\)\n- \\(7^{16} = (7^8)^2 = 51^2 = 2601 \\equiv 2601 - 20 \\times 125 = 2601 - 2500 = 101 \\pmod{125}\\)\n\nNow, express \\(26\\) in binary: \\(26 = 16 + 8 + 2\\). Thus,\n\\[\n7^{26} = 7^{16} \\times 7^8 \\times 7^2 \\equiv 101 \\times 51 \\times 49 \\pmod{125}\n\\]\n\nFirst, compute \\(101 \\times 51\\):\n- \\(100 \\times 51 = 5100\\)\n- \\(1 \\times 51 = 51\\)\n- Total: \\(5100 + 51 = 5151\\)\n\nNow reduce modulo 125:\n- \\(125 \\times 41 = 5125\\)\n- \\(5151 - 5125 = 26\\), so \\(101 \\times 51 \\equiv 26 \\pmod{125}\\)\n\nNext, multiply by 49:\n\\[\n26 \\times 49 = 1274\n\\]\nReduce modulo 125:\n- \\(125 \\times 10 = 1250\\)\n- \\(1274 - 1250 = 24\\), so \\(26 \\times 49 \\equiv 24 \\pmod{125}\\)\n\nThus,\n\\[\n7^{26} \\equiv 24 \\pmod{125}\n\\]\n\nSo, we have:\n\\[\n7^{2026} \\equiv 1 \\pmod{8} \\quad \\text{and} \\quad 7^{2026} \\equiv 24 \\pmod{125}\n\\]\n\n**Step 3: Apply Chinese Remainder Theorem**\n\nWe need to find a number \\(x\\) such that:\n\\[\n\\begin{cases}\nx \\equiv 1 \\pmod{8} \\\\\nx \\equiv 24 \\pmod{125}\n\\end{cases}\n\\]\n\nLet \\(x = 125k + 24\\) for some integer \\(k\\). Substitute into the first congruence:\n\\[\n125k + 24 \\equiv 1 \\pmod{8}\n\\]\nCompute \\(125 \\mod 8\\):\n- \\(125 \\div 8 = 15 \\times 8 = 120\\), remainder 5, so \\(125 \\equiv 5 \\pmod{8}\\)\nThus,\n\\[\n5k + 24 \\equiv 1 \\pmod{8} \\implies 5k + 0 \\equiv 1 \\pmod{8} \\quad (\\text{since } 24 \\div 8 = 3)\n\\]\nSo,\n\\[\n5k \\equiv 1 \\pmod{8}\n\\]\n\nSolve for \\(k\\): Multiply both sides by the modular inverse of 5 modulo 8. Since \\(5 \\times", "_no tool calls_"], "python": ["**Final answer:** `649` \n**Generated tokens:** 281 \u00b7 **Time:** 19.8 s \u00b7 **Model calls:** 2 \u00b7 **Tool calls:** 1", "We need the remainder when \\(7^{2026}\\) is divided by 1000. Computing directly with modular exponentiation gives:\n\n\\[\n7^{2026} \\bmod 1000 = 649.\n\\]\n\nThus, the remainder is \\(\\boxed{649}\\).", "**run_python #1**\n```python\nprint(pow(7, 2026, 1000))\n\n```\n```text\n649\n```"]}, {"q": "What is 987654321 \u00d7 123456789?", "gold": "121932631112635269", "none": ["**Final answer:** `(no boxed answer)` \n**Generated tokens:** 1,024 \u00b7 **Time:** 61.7 s \u00b7 **Model calls:** 1 \u00b7 **Tool calls:** 0 \n\u26a0\ufe0f hit the token cap", "We need to compute the product \\(987,654,321 \\times 123,456,789\\). \n\nNotice that these numbers are related: \n\\(987,654,321 = 10^9 - 12,345,679\\)? Not exactly. Alternatively, observe that:\n\\[\n987,654,321 = 1,000,000,000 - 12,345,679\n\\]\nBut maybe there's a pattern with the digits. Another approach: \nLet \\(a = 123,456,789\\). Then note that \\(987,654,321\\) is like reversing the digits of \\(123,456,789\\)? Actually, \\(123,456,789\\) reversed is \\(987,654,321\\). Yes! Because:\n- Original: 1 2 3 4 5 6 7 8 9\n- Reversed: 9 8 7 6 5 4 3 2 1 = 987,654,321.\n\nSo we need to compute \\(a \\times \\text{reverse}(a)\\) where \\(a = 123,456,789\\). There might be a known identity or we can compute directly using multiplication properties. \n\nWe can use the formula for multiplying a number by its reverse? Alternatively, just do standard multiplication but maybe there is a trick: \nLet \\(N = 123,456,789\\). Then \\(R(N) = 987,654,321\\). We want \\(N \\times R(N)\\). \n\nOne way: Write both numbers in terms of sums. But perhaps easier to compute using Python-like mental arithmetic? Since it's a contest problem maybe there is a neat result.\n\nLet me attempt to compute step by step:\n\nWe can use the identity:\n\\[\n(10^9 - 123,456,789) \\times (10^8 + 1,234,567,89?) \\] Not sure.\n\nAlternatively, note that \\(987,654,321 = 1000000000 - 12,345,679\\). But 12,345,679 is related to 123,456,789? Actually, 123,456,789 minus something? Let's check: \n\\(123,456,789 - 9,111,110 = ?\\) Not obvious.\n\nMaybe we can compute using the fact that:\n\\[\n987,654,321 \\times 123,456,789 = (1,000,000,000 - 12,345,679) \\times 123,456,789\n\\]\nBut then \\(12,345,679\\) is not directly related to 123,456,789.\n\nAnother idea: Use the property that for any number \\(x\\), \n\\(x \\times (10^9 - x) = 10^9 x - x^2\\). But here we have reverse, not complement.\n\nLet's just do direct multiplication using some systematic approach. Since numbers are 9-digit, product will be up to 18 digits. We can compute by splitting into parts or use known patterns: \n\\(123,456,789 \\times 987,654,321\\) is symmetric? Possibly the result might have a pattern like 121... but let's not guess.\n\nI'll attempt to compute using long multiplication with some shortcuts. Alternatively, I can break down:\n\nLet \\(A = 123,456,789\\). Write it as:\n\\(A = 123,000,000 + 456,789\\)? Not great.\n\nBetter: Use the fact that both numbers are close to powers of 10 and each other. But maybe we can compute using a calculator in mind? That's error-prone. I'll try to do step-by-step multiplication with grouping.\n\nWe want \\(A \\times B\\), where \\(B = 987,654,321\\). Notice that \\(B = 1,000,000,000 - 12,345,679\\) as earlier. So:\n\\[\nA \\times B = A \\times (10^9 - 12,345,679) = A \\times 10^9 - A \\times 12,345,679.\n\\]\nNow \\(A \\times 10^9 = 123,456,789,000,000,000\\)? Actually, \\(A = 123,456,789\\). Multiply by \\(10^9\\) gives \\(123,456,789,000,000,000\\) (since appending nine zeros). That's a huge number: 123,456,789 followed by 9 zeros.\n\nThen subtract \\(A \\times 12,345,679\\). So we need to compute \\(123,456,789 \\times 12,345,679\\). That might be simpler? Possibly not.\n\nAlternatively, use the reverse property: There is known identity for numbers like 123...n and its reverse. For n=9, maybe there's a pattern. But let's try direct multiplication with some help from known sums.\n\nI can compute using Python", "_no tool calls_"], "python": ["**Final answer:** `121932631112635269` \n**Generated tokens:** 151 \u00b7 **Time:** 11.1 s \u00b7 **Model calls:** 2 \u00b7 **Tool calls:** 1", "We need to compute \\(987,654,321 \\times 123,456,789\\). Using Python's arbitrary-precision integer arithmetic:\n\n```python\n987654321 * 123456789\n```\n\nThe result is `121932631112635269`.\n\nThus, the product is \\(\\boxed{121932631112635269}\\).", "**run_python #1**\n```python\n987654321 * 123456789\n```\n```text\n121932631112635269\n```"]}, {"q": "How many ways can you make change for one dollar using pennies, nickels, dimes and quarters?", "gold": "242", "none": ["**Final answer:** `(no boxed answer)` \n**Generated tokens:** 1,024 \u00b7 **Time:** 59.4 s \u00b7 **Model calls:** 1 \u00b7 **Tool calls:** 0 \n\u26a0\ufe0f hit the token cap", "We need to find the number of ways to make change for one dollar (100 cents) using pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents). This is a classic coin change problem where we count the combinations of these coins that sum to 100 cents.\n\nLet:\n- \\( p \\) = number of pennies (0 or more)\n- \\( n \\) = number of nickels (0 or more)\n- \\( d \\) = number of dimes (0 or more)\n- \\( q \\) = number of quarters (0 or more)\n\nWe have the equation:\n\\[ 1p + 5n + 10d + 25q = 100 \\]\n\nWe can approach this by fixing the number of quarters (\\( q \\)), then for each, fix the number of dimes (\\( d \\)), and then count the possibilities for nickels and pennies.\n\nAlternatively, we can iterate over possible values of \\( q \\) from 0 to 4 (since \\( 25 \\times 4 = 100 \\)). For each \\( q \\), let the remaining amount be \\( R_q = 100 - 25q \\). Then we need to make \\( R_q \\) cents using nickels, dimes, and pennies.\n\nFor a fixed \\( q \\), we can iterate over possible \\( d \\) from 0 to \\( \\lfloor R_q / 10 \\rfloor \\). For each \\( d \\), let the remaining amount be \\( R_{qd} = R_q - 10d \\). Then we need to make \\( R_{qd} \\) cents using nickels and pennies. Since pennies can fill any remainder, for a given \\( n \\) (number of nickels), the number of pennies is determined as \\( p = R_{qd} - 5n \\), which must be non-negative. So for fixed \\( q \\) and \\( d \\), the number of solutions in \\( n \\) is the number of integers \\( n \\) such that:\n\\[ 0 \\leq n \\leq \\left\\lfloor \\frac{R_{qd}}{5} \\right\\rfloor \\]\nand also \\( p = R_{qd} - 5n \\geq 0 \\). Since \\( p \\) can be any non-negative integer, as long as \\( n \\leq \\lfloor R_{qd}/5 \\rfloor \\), it's valid. So the number of choices for \\( n \\) is \\( \\left\\lfloor \\frac{R_{qd}}{5} \\right\\rfloor + 1 \\).\n\nThus, for each \\( q \\) and \\( d \\), the number of ways to choose nickels and pennies is \\( \\lfloor R_{qd}/5 \\rfloor + 1 \\).\n\nNow we compute:\n\n**Case \\( q = 0 \\):**\n- \\( R_0 = 100 \\)\n- \\( d \\) from 0 to \\( \\lfloor 100/10 \\rfloor = 10 \\):\n - For each \\( d \\), \\( R_{qd} = 100 - 10d \\).\n - Number of nickels choices: \\( \\left\\lfloor \\frac{100 - 10d}{5} \\right\\rfloor + 1 = (20 - 2d) + 1? \\) Let's compute carefully:\n \\[ \\frac{100 - 10d}{5} = 20 - 2d \\]\n So floor is \\( 20 - 2d \\) (since it's integer). Then number of choices for \\( n \\) is \\( (20 - 2d) + 1 = 21 - 2d \\).\n - But we must ensure that \\( d \\leq 10 \\), and when \\( d=10 \\), \\( R_{qd}=0 \\), then \\( \\lfloor 0/5 \\rfloor = 0 \\), so choices = 0+1=1. Our formula gives \\( 21 - 2*10 = 1 \\), correct.\n - So for each \\( d \\) from 0 to 10, number of ways = \\( 21 - 2d \\).\n\nLet's sum over \\( d \\):\n\\[ \\sum_{d=0}^{10} (21 - 2d) \\]\nThis is an arithmetic series. Number of terms: 11.\nFirst term (\\( d=0 \\)): 21\nLast term (\\( d=10 \\)): 1\nSum = number of terms * (first + last)/2 = \\( 11 \\times (21+1)/2 = 11 \\times 22/2 = 11 \\times 11 = 121 \\).\n\n**Case \\( q = 1 \\):**\n- \\( R_1 = 100 - 25 = 75 \\)\n- \\( d \\) from 0 to \\( \\lfloor 75/10 \\rfloor = 7 \\):\n For each \\( d \\), \\( R_{qd} = 75 - 10d \\).\n Number of nickels choices: \\( \\left\\lfloor \\frac{75 - 10d}{5} \\right\\rfloor + 1 = (15 - 2d) + 1? \\) Compute:\n \\[ \\frac{75 - 10d}{5", "_no tool calls_"], "python": ["**Final answer:** `242` \n**Generated tokens:** 783 \u00b7 **Time:** 57.2 s \u00b7 **Model calls:** 3 \u00b7 **Tool calls:** 2", "The code counts all nonnegative integers \\(q\\) (quarters), \\(d\\) (dimes), and \\(n\\) (nickels) such that the total value of those coins does not exceed 100 cents. For each such triple, the remaining amount is made up by pennies, giving exactly one way to form a dollar. The loops over quarters from 0 to 4, dimes from 0 to 10, and nickels from 0 to 20 cover all possibilities, and the condition \\(25q + 10d + 5n \\le 100\\) yields 242 valid triples.\n\nThus there are **242** different ways to make change for one dollar using pennies, nickels, dimes, and quarters.\n\n\\[\n\\boxed{242}\n\\]", "**run_python #1**\n```python\ncount = 0\nfor q in range(0, 5): # quarters: 0 to 4 (since 5*25=125 >100)\n for d in range(0, 11): # dimes: 0 to floor((100-25q)/10)+1? Let's just go up to 10\n for n in range(0, 21): # nickels: 0 to floor((100-25q-10d)/5)+1? We'll bound by 20\n if 25*q + 10*d + 5*n <= 100:\n count += 1\nprint(count)\n\n```\n```text\n242\n```\n\n**run_python #2**\n```python\n# Let's verify by generating all combinations via loops and counting distinct sums? Actually we counted each triple (q,d,n) that satisfies inequality, which corresponds to a unique way including pennies. That should be correct.\n\n```\n```text\n(no output)\n```"]}, {"q": "How many primes are there below 100,000?", "gold": "9592", "none": ["**Final answer:** `9592` \n**Generated tokens:** 217 \u00b7 **Time:** 12.7 s \u00b7 **Model calls:** 1 \u00b7 **Tool calls:** 0", "To determine the number of prime numbers below 100,000, we can use the Prime Number Theorem or known results from prime counting functions. The exact count is given by \\(\\pi(10^5)\\), where \\(\\pi(x)\\) denotes the number of primes less than or equal to \\(x\\).\n\nFrom standard tables or computational results:\n- \\(\\pi(10^4) = 1229\\) (primes below 10,000)\n- \\(\\pi(10^5) = 9592\\)\n\nHowever, let's verify this more accurately. The Prime Number Theorem states that \\(\\pi(x) \\approx \\frac{x}{\\ln x}\\), but for exact counts up to 100,000, we can rely on known values.\n\nUpon checking reliable sources or calculating:\n- The number of primes below 10,000 is indeed 1229.\n- For 100,000, the count is 9592. This is a well-established result in number theory.\n\nThus, there are **9592** prime numbers below 100,000.\n\n\\[\n\\boxed{9592}\n\\]", "_no tool calls_"], "python": ["**Final answer:** `9592` \n**Generated tokens:** 526 \u00b7 **Time:** 34.9 s \u00b7 **Model calls:** 2 \u00b7 **Tool calls:** 1", "The Python code computed the number of primes below 100,000 using the Sieve of Eratosthenes. The result is **9592**.\n\nThus, there are \\(\\boxed{9592}\\) primes below 100,000.", "**run_python #1**\n```python\ndef count_primes(limit):\n sieve = [True] * (limit+1)\n sieve[0] = sieve[1] = False\n for i in range(2, int(limit**0.5) + 1):\n if sieve[i]:\n step = i\n start = i*i\n # mark multiples of i starting from i*i as False\n for j in range(start, limit+1, step):\n sieve[j] = False\n return sum(sieve)\n\nprint(count_primes(100000))\n\n```\n```text\n9592\n```"]}]; const REC="### Recorded results\n\nThe main win is speed: fewer generated tokens at about the same accuracy. The tool did not raise accuracy once the model had its full context. The AIME gains in the October 28 runs further down came from the token cap, because the no-tools side ran out of tokens more often.\n\n### Full-context results (Oct 30, 2026)\n\nModel: full-precision (BF16) LFM3-1.7B on a Hugging Face A100 with SGLang, 262,144-token context, the model card's sampling (temperature 1.0, top_p 0.95, top_k 20), one sample per problem, graded with `math_verify`. Up to 6 `run_python` calls, 10 s timeout. No answer ran out of context.\n\n| Set | No tools | run_python | run_python + check |\n|---|---|---|---|\n| AIME 2025 (30) | 24/30 | 22/30 | 25/30 |\n| AIME 2026 (30) | 26/30 | 25/30 | 25/30 |\n| MATH-500 (500) | 486/500 | 486/500 | 481/500 |\n| AIME tokens per solved problem | 21,776 | 16,688 (\u221223.4%) | 17,495 (\u221219.7%) |\n| MATH-500 tokens per solved problem | 3,752 | 3,221 (\u221214.2%) | 3,510 (\u22126.4%) |\n\nMcNemar exact, two-sided: on all 60 AIME problems, run_python vs no tools fixed 3 and broke 6 (p = 0.51); run_python + check vs no tools fixed 5 and broke 5 (p = 1.0). On MATH-500 the p-values were 1.0 and 0.30. None of the accuracy differences is significant.\n\nThese cloud runs did not use the tool's network-denied sandbox. The Python code ran as the `nobody` user, and the network probe in each job's meta file returned `NET_OK`, so the code had network access there.\n\nWrite-up: https://www.destroyrebuild.xyz/blog/lfm3-python-tool-full-context\n\n### Earlier capped runs (Oct 28, 2026)\n\nThese ran with a 25,000-token budget per problem. The accuracy changes below were cap effects and did not hold at full context. The token savings did.\n\n| Set | Model | No tools | run_python | Tokens / problem | Change |\n|---|---|---|---|---|---|\n| MATH (30) | LFM3-1.7B | 30/30 | 29/30 | 3,036.8 \u2192 2,022.8 | \u221233.4% |\n| MATH (30) | Clover-35B-A3B | 28/30 | 28/30 | 2,791.7 \u2192 1,458.6 | \u221247.8% |\n| AIME 2025 (30) | LFM3-1.7B | 17/30 | 20/30 | 16,634.1 \u2192 9,595.6 | \u221242.3% |\n| MATH-500 L5 (60) | LFM3-1.7B | 54/60 | 50/60 | 7,577.1 \u2192 5,971.6 | \u221221.2% |\n| Combined (90) | LFM3-1.7B | 71/90 | 70/90 | 10,596.1 \u2192 7,179.6 | \u221232.2% |\n\nTime per problem on MATH (30): LFM3 24.18 s \u2192 16.64 s (\u221231.2%). Clover 131.68 s \u2192 69.44 s (\u221247.3%). I only quote tokens for AIME and L5, because part of the no-tools side ran with parallel workers on a shared server.\n\nAccuracy (Wilson 95% CI, McNemar exact p):\n\n- LFM3 on MATH: 30/30 (88.6\u2013100%) without tools. 29/30 with Python by the grader, 30/30 by hand. One answer was a formatting miss.\n- Clover on MATH: 28/30 both ways, p = 1.0.\n- AIME 2025: 17/30 (39.2\u201372.6%) \u2192 20/30 (48.8\u201380.8%), +10.0 pts. Fixed 4, broke 1, p = 0.375. Not significant. Token-cap hits 11 \u2192 2.\n- MATH-500 L5: 54/60 (79.9\u201395.3%) \u2192 50/60 (72.0\u201390.7%), \u22126.7 pts. Fixed 0, broke 4, p = 0.125.\n- Combined 90: fixed 4, broke 5, p = 1.0. Token-cap hits 14 \u2192 3.\n- `--verify` variant on AIME: 17/30 \u2192 24/30 (62.7\u201390.5%), +23.3 pts, fixed 8, broke 1, p = 0.039. Tokens \u221229.8%. This is preliminary and post-hoc. I tried it after seeing the base results, and its L5 run had only 3 of 60 done at publish time.\n\nSetup: llama.cpp `llama-server`, GGUF, greedy, thinking on, 25,000 generated-token budget per problem summed across all model calls. Up to 6 `run_python` calls, 10 s timeout per call, output capped at 2,000 characters. Apple M1 Max.\n\nRaw data, the benchmark script and the grader: https://huggingface.co/austindixson/python-sandbox-math-tool\n\n### Demo runs on 2 CPU threads (same app, 1,024-token cap)\n\nRecorded before publishing, on a 2-thread Linux machine (bwrap sandbox) with this Space's own `live/app.py`. These are demo-setting numbers, not the eval above.\n\n| Question | Correct answer | No tools | run_python |\n|---|---|---|---|\n| What is the sum of the digits of 2^1000? | 1366 | (no boxed answer) \u00b7 1,024 tok \u00b7 61.5 s | 1366 \u00b7 153 tok \u00b7 12.3 s |\n| What is the remainder when 7^2026 is divided by 1000? | 649 | (no boxed answer) \u00b7 1,024 tok \u00b7 61.4 s | 649 \u00b7 281 tok \u00b7 19.8 s |\n| What is 987654321 \u00d7 123456789? | 121932631112635269 | (no boxed answer) \u00b7 1,024 tok \u00b7 61.7 s | 121932631112635269 \u00b7 151 tok \u00b7 11.1 s |\n| How many ways can you make change for one dollar using pennies, nickels, dimes and quarters? | 242 | (no boxed answer) \u00b7 1,024 tok \u00b7 59.4 s | 242 \u00b7 783 tok \u00b7 57.2 s |\n| How many primes are there below 100,000? | 9592 | 9592 \u00b7 217 tok \u00b7 12.7 s | 9592 \u00b7 526 tok \u00b7 34.9 s |\n\nA sixth question I tried (integers up to 10,000 divisible by 3 or 7 but not 5) hit the 1,024-token cap both ways without calling the tool, so I left it out of the presets.\n"; |
