Datasets:
id stringlengths 68 68 | source stringclasses 2
values | question stringlengths 14 1.22k | steps listlengths 3 45 | answer stringlengths 15 3.01k |
|---|---|---|---|---|
nemotron-math-v2-medium-0000104-dd0a15df-3fed-6f9b-3534-bc2b6f1f4e5e | nvidia/Nemotron-Math-v2/aops | For how many consecutive integers can the Möbius function \( \mu(n) \) take the value 0? | [
"We need to determine the maximum length of a run of consecutive integers n where μ(n)=0. μ(n)=0 iff n is not squarefree, i.e., divisible by a squared prime. So we need longest possible consecutive sequence of integers each divisible by a square of a prime. Find maximum length of a block of consecutive integers eac... | The Möbius function
\[
\mu(n)=\begin{cases}
0 &\text{if }n\text{ is not square‑free (i.e., }p^{2}\mid n\text{ for some prime }p),\\[2mm]
(-1)^{\omega(n)} &\text{if }n\text{ is square‑free},
\end{cases}
\]
so \(\mu(n)=0\) exactly when \(n\) is divisible by a square of a prime.
Thus a run of consecutive integers
... |
nemotron-math-v2-medium-0000242-e3b8c491-897d-953e-8594-9faf3186e301 | nvidia/Nemotron-Math-v2/aops | In how many ways may 5 people be seated in a row of twenty chairs given that no two people may sit next to one another? | [
"We have 20 chairs in a row, seat 5 people such that no two are adjacent. We need count of ways: choose chairs positions (with gaps) then arrange people (5! permutations). Classic stars-and-bars. Choose 5 chairs out of 20 with no two adjacent. Number of ways to choose k positions from n such that no two adjacent is... | To count the arrangements we separate the problem into two parts:
1. **Choose which chairs will be occupied** (the positions must be non‑adjacent).
2. **Assign the 5 distinct people to those chosen chairs**.
---
### 1. Choosing chairs with no two adjacent
Let the 20 chairs be numbered \(1,2,\dots ,20\).
We need... |
nemotron-math-v2-medium-0000524-9e386a02-a3cc-4a5b-1775-2ac6b3f2b8bf | nvidia/Nemotron-Math-v2/aops | Find all positive integers \( k \) for which the equation \( n^m - m^n = k \) has a solution in positive integers. | [
"We need all positive integers k such that there exist positive integers n,m with n^m - m^n = k. We need characterize possible values. Let's explore. This is similar to difference of exponentials. For any k we can try to find n,m? Might be all integers except maybe 0? Need positive k. Let's test small values.",
"... | Let
\[
n^{m}-m^{n}=k ,\qquad n,m\in\mathbb Z_{>0},
\]
and ask for which positive integers \(k\) such an equation can be satisfied.
--------------------------------------------------------------------
### 1. A simple construction
Take the exponent of the second term equal to \(1\); i.e. put
\[
m=1 .
\]
Then ... |
nemotron-math-v2-medium-0000595-ce871de5-da11-7dab-3917-dc22a5f7858c | nvidia/Nemotron-Math-v2/aops | How many distinct integer factor pairs of $10500$ are there such that their product is $10500$? | [
"We need number of distinct integer factor pairs (a,b) such that a*b = 10500. Distinct integer factor pairs presumably unordered? Usually factor pair means unordered pair (a,b) with a ≤ b. Also can be negative? Usually integer factor pairs include negative as well? Typically factor pairs are positive. Problem says ... | The prime factorisation of \(10500\) is
\[
10500 = 2^{2}\cdot 3^{1}\cdot 5^{3}\cdot 7^{1}.
\]
---
### 1. Count the positive divisors
For a number \(n = p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\) the number of positive divisors is
\[
(e_1+1)(e_2+1)\cdots (e_k+1).
\]
Hence
\[
\tau(10500) = (2+1)(1+1)(3+1)(1+1)=3... |
nemotron-math-v2-medium-0000652-fb162d81-f8c8-77cc-4520-50b5d2a208c8 | nvidia/Nemotron-Math-v2/aops | What is the sixth perfect number, and does it end in 28? | [
"The user asks: \"What is the sixth perfect number, and does it end in 28? Please reason step by step, and put your final answer within \\boxed{}.\"",
"We need to compute perfect numbers: They are numbers equal to sum of proper divisors. Known even perfect numbers are given by Euclid-Euler: 2^{p-1}(2^p -1) where ... | **Step‑by‑step reasoning**
1. **Even perfect numbers** are given by Euclid’s formula
\[
N = 2^{p-1}\bigl(2^{p}-1\bigr)
\]
where \(2^{p}-1\) is a Mersenne prime.
2. The known exponents \(p\) that make \(2^{p}-1\) prime (in increasing order) are
\[
p = 2,\,3,\,5,\,7,\,13,\,17,\,19,\dots
\]
3.... |
nemotron-math-v2-medium-0000872-177a8227-355b-75d1-205f-c77690b80e2b | nvidia/Nemotron-Math-v2/aops | Given a clock with two hands at the 3 o'clock and 10 o'clock markers, what is the area of the smaller region of the circle formed by the angle between the two hands? | [
"We have a clock: two hands at 3 o'clock and 10 o'clock markers. So positions: 3 o'clock is at the 3 position, which is at 90 degrees from 12 (i.e., to the right horizontally). 10 o'clock marker is at 10, which is 10/12 of full circle? Actually each hour marker is 30 degrees apart. Starting at 12 (0°), moving clock... | **Step 1: Convert the clock‑hand positions to angles.**
On an analog clock each hour marker is \(30^{\circ}\) apart (\(360^{\circ}/12\)).
- The hand at **3 o’clock** points \(3\times30^{\circ}=90^{\circ}\) clockwise from the 12‑o’clock position.
- The hand at **10 o’clock** points \(10\times30^{\circ}=300^{\circ... |
nemotron-math-v2-medium-0001059-5a31440e-8ec6-3c16-dd6b-779768a1ddfd | nvidia/Nemotron-Math-v2/aops | What is the closest average interplanetary speed in miles per hour of NASA’s Perseverance Rover, which traveled $292{,}526{,}838$ miles over a period of approximately $6.5$ months? | [
"We need average speed = total distance / time. Distance = 292,526,838 miles. Time approx 6.5 months. Convert months to hours. 1 month approximated as 30.44 days? But maybe use 30 days? Usually average month ~30.44 days (365.25/12). Let's compute both, then decide closest. Use 6.5 months * 30.44 days/month = 197.86... | The average speed is the total distance divided by the total time.
**1. Convert the time to hours**
A month is about \(30.44\) days (the average length of a month in a year).
\[
6.5 \text{ months}=6.5 \times 30.44\text{ days}=197.86\text{ days}
\]
\[
197.86\text{ days}\times 24\text{ h/day}=4\,748.6\text{ h}
\]
**... |
nemotron-math-v2-medium-0001222-2333bc6e-b59d-08df-5563-e5b7166194aa | nvidia/Nemotron-Math-v2/aops | Evaluate the sum \(\sum_{k=2, k \text{ even}}^{\infty} \frac{1}{k!}\). | [
"We need to evaluate sum over k from 2 to infinity, only even k, of 1/k!. So sum_{n=1 to ∞} 1/(2n)!? Starting k=2 (even) then 4,6,... So sum_{n=1}^\\infty 1/(2n)!.",
"We can express using series for e^x: e^x = sum_{k=0}∞ x^k / k!. For x=1, e = sum_{k=0}∞ 1/k!. Even and odd parts: cosh(1) = sum_{n=0}∞ 1/(2n)!, sin... | The series contains only the even‑indexed terms of the exponential series:
\[
\sum_{k=2,\,k\text{ even}}^{\infty}\frac1{k!}
=\sum_{n=1}^{\infty}\frac1{(2n)!}.
\]
Recall the Taylor expansions
\[
e^{x}= \sum_{k=0}^{\infty}\frac{x^{k}}{k!},\qquad
\cosh x = \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}.
\]
Setting \(x=1\)... |
nemotron-math-v2-medium-0001235-b8218f1a-5dc7-90e0-0d7b-637cee2ded3f | nvidia/Nemotron-Math-v2/aops | Given the quantities $a+b$, $ab$, and $b^2$, express $b$ using only arithmetic operations (addition, subtraction, multiplication, division). | [
"We need to express b in terms of given quantities: S = a+b, P = ab, and Q = b^2. So we have S = a+b, P = a b, Q = b^2. Want b expressed solely using S, P, Q with arithmetic operations.",
"We can treat unknowns a, b satisfy these equations. Solve for b.",
"We have Q = b^2 => b = ± sqrt(Q). But sqrt is not allow... | Let
\[
S = a+b,\qquad P = ab,\qquad Q = b^{2}.
\]
From \(S=a+b\) we have \(a = S-b\).
Substituting this into \(P=ab\):
\[
P = (S-b)\,b = Sb - b^{2}=Sb-Q .
\]
Hence
\[
Sb = P+Q \quad\Longrightarrow\quad b = \frac{P+Q}{S}.
\]
Returning to the original symbols, the expression for \(b\) using only the given qua... |
nemotron-math-v2-medium-0001383-4db9ee59-793a-5ed9-756b-f7ed022628ae | nvidia/Nemotron-Math-v2/aops | A landscape gardener uses 36 paving bricks, each measuring 22cm by 11cm, to form a 'circular' flower bed. What is the radius of the circle? | [
"We need to interpret problem: a landscape gardener uses 36 paving bricks, each measuring 22 cm by 11 cm, to form a 'circular' flower bed. What is the radius of the circle? Likely the bricks are arranged edge-to-edge approximating a circle, perhaps laying them radially like a circular arrangement perimeter? Or perh... | The 36 bricks are laid side‑by‑side around the edge of the flower bed, so their long sides form the perimeter of the circle.
\[
\text{Circumference}=36 \times 22\text{ cm}=792\text{ cm}
\]
For a circle,
\[
C = 2\pi r \quad\Longrightarrow\quad r=\frac{C}{2\pi}
=\frac{792}{2\pi}
=\frac{396}{\pi}\text{ cm}
\... |
nemotron-math-v2-medium-0001457-3830b1bc-b14e-6a33-ac72-28b1fae08611 | nvidia/Nemotron-Math-v2/aops | Evaluate the sum \( S = \sum_{k=1}^n \sin^2(nx) \), where \( n \in \mathbb{N}^* \). | [
"We need to evaluate S = \\sum_{k=1}^n sin^2(nx). Wait variable k doesn't appear inside? It says sum from k=1 to n sin^2(nx). That seems independent of k, so sum = n * sin^2(nx). Probably a typo: maybe sum_{k=1}^n sin^2(kx). Usually interesting sum is sum_{k=1}^n sin^2(kx). Let's assume that. So S = Σ_{k=1}^n sin^2... | \[
S=\sum_{k=1}^{n}\sin ^2(kx)
\]
---
**Step 1. Use the double‑angle identity**
\[
\sin ^2(kx)=\frac{1-\cos (2kx)}{2}.
\]
Hence
\[
S=\sum_{k=1}^{n}\frac{1-\cos (2kx)}{2}
=\frac{n}{2}-\frac12\sum_{k=1}^{n}\cos (2kx).
\]
---
**Step 2. Sum the cosine progression**
For a constant angle \(\theta\),
\[
\sum_{... |
nemotron-math-v2-medium-0001934-8d4aa151-cece-abbe-87ea-d854ed6b4fff | nvidia/Nemotron-Math-v2/aops | For a prime number \( p > 2 \), how many integers \( x \) in the set \( \{1, 2, \ldots, p-1\} \) satisfy \( x^3 \equiv 1 \pmod{p} \)? | [
"We need number of solutions to x^3 ≡1 mod p with p prime >2. In multiplicative group of order p-1 (cyclic). Number of solutions equals gcd(3, p-1). Because equation a^k=1 in cyclic group of order n has exactly gcd(k,n). So answer = gcd(3, p-1). Since p>2 prime, p-1 even. Cases: if p ≡1 mod 3 then gcd=3. If p ≡2 mo... | In the multiplicative group
\[
(\mathbb Z/p\mathbb Z)^{\times}=\{1,2,\dots ,p-1\},
\]
the operation is multiplication modulo \(p\).
For a prime \(p>2\) this group is **cyclic** of order \(p-1\).
---
### Step 1: Reduce the problem to an exponent equation
Let \(g\) be a generator (primitive root) of \((\mathbb Z... |
nemotron-math-v2-medium-0001948-615c2dd2-cd1c-e4b3-893f-f6e9f4306201 | nvidia/Nemotron-Math-v2/aops | Given \( S = \frac{n(n+1)(2n+1)}{6} \) (the sum of squares from 1 to \( n \)), define \( S' = S - 1 \). If \( n \) is not a perfect square and \( S' \mod n = n - 1 \), then \( n \) is prime. Verify this for \( n = 71 \). | [
"We need to verify claim: given S = n(n+1)(2n+1)/6 sum of squares. Define S' = S - 1. If n is not a perfect square and S' mod n = n-1 then n is prime. For n=71, check conditions.",
"Compute S = 71*72*(2*71+1)/6 = 71*72*143/6. Simplify: 72/6 =12. So S = 71*12*143 = (71*12)*143 = 852*143? Wait 71*12 = 852. Then 852... | **Step 1: Compute the sum of squares \(S\).**
\[
S=\frac{n(n+1)(2n+1)}{6}
=\frac{71\cdot72\cdot(2\cdot71+1)}{6}
=\frac{71\cdot72\cdot143}{6}
\]
Since \(\dfrac{72}{6}=12\),
\[
S =71\cdot12\cdot143 = 852\cdot143 = 121\,836 .
\]
---
**Step 2: Form \(S' = S-1\).**
\[
S' = 121\,836 - 1 = 121\,835 .
\]
-... |
nemotron-math-v2-medium-0001981-85c436c4-1543-e76d-0a5e-cb1a951fb00c | nvidia/Nemotron-Math-v2/aops | Using sigma notation, find the number and the sum of its digits for a number in base \( q \) represented by \( a_{n}a_{n-1}a_{n-2}...a_{3}a_{2}a_{1}a_{0} \). | [
"We need to interpret the problem. \"Using sigma notation, find the number and the sum of its digits for a number in base q represented by a_n a_{n-1} ... a_0.\" Probably they want expression for the value of the number (the integer) using sigma notation, as well as expression for sum of its digits. So the \"number... | Let
\[
N = a_{n}a_{n-1}a_{n-2}\dots a_{3}a_{2}a_{1}a_{0}
\]
be a numeral written in base \(q\) (with \(q\ge 2\)).
The right‑most digit is \(a_{0}\) (the units digit), the next digit is \(a_{1}\) (the \(q\)’s place), and so on, up to the left‑most digit \(a_{n}\) (the \(q^{n}\)‑place).
---
### 1. Value of the n... |
nemotron-math-v2-medium-0002038-e9aa91d6-7f64-890b-fd72-92de911042a6 | nvidia/Nemotron-Math-v2/aops | Given a clock with two hands at the 3 o'clock and 10 o'clock markers, what is the area of the smaller region of the circle formed by the angle between the two hands? | [
"We have a clock face, a circle. Hands at 3 o'clock and 10 o'clock markers. Need area of smaller region formed by the angle between the two hands. So we consider the angle between the rays from center to 3 and 10. Determine the smaller central angle, then compute area of sector = (θ/2π) * πr^2 = (θ/2) * r^2? Actual... | The hands point to the hour‑marks “3” and “10”.
Each hour‑mark is \(30^{\circ}\) apart, so measured from the 12‑o’clock position we have
\[
\begin{aligned}
\text{3 o’clock} &: 3 \times 30^{\circ}=90^{\circ},\\
\text{10 o’clock}&:10 \times 30^{\circ}=300^{\circ}.
\end{aligned}
\]
The absolute difference of these d... |
nemotron-math-v2-medium-0002277-ee85a511-cd3f-2937-3527-78687c14d4ad | nvidia/Nemotron-Math-v2/aops | Given the functional equation \( f(xy) = f(x) + f(y) \) for positive real numbers \( x \) and \( y \), determine an expression for \( f \) involving one variable and two parameters. | [
"We need to solve Cauchy-type functional equation on positive reals: f(xy)=f(x)+f(y). This is the logarithm property, solutions are f(x)=k ln x for some constant k, assuming continuity or measurability etc. But the problem: \"determine an expression for f involving one variable and two parameters.\" They want gener... | The functional equation
\[
f(xy)=f(x)+f(y)\qquad (x>0,\;y>0)
\]
is the multiplicative version of Cauchy’s additive equation.
Take a logarithm of the arguments: let
\[
g(t)=f(e^{t}), \qquad t\in\mathbb {R}.
\]
Since \(e^{t_1}e^{t_2}=e^{t_1+t_2}\),
\[
g(t_1+t_2)=f(e^{t_1+t_2})=f(e^{t_1}e^{t_2})
=f(e^... |
nemotron-math-v2-medium-0002388-e63fee69-80a7-94bb-ae51-91ea154559d3 | nvidia/Nemotron-Math-v2/aops | Let \( t_n \) denote the number of binary sequences of length \( n \) that do not contain three adjacent 1s. Determine a recurrence relation for \( t_n \), and provide enough initial values to generate the sequence. | [
"We need recurrence for number of binary strings length n without \"111\". Classic linear recurrence: t_n = t_{n-1} + t_{n-2} + t_{n-3}. Explanation: Consider first block? Let's derive. For n ≥ 3, count strings ending with 0,10,110. But standard: Let a_n count strings length n with no three consecutive 1's. Then a_... | Let
\[
t_n=\#\{\text{binary strings of length }n\text{ with no three consecutive }1\text{s}\}.
\]
---
### Step 1 – Classify the strings by their last block
A binary string that avoids “111’’ can end in three possible ways:
1. **…0** – the last symbol is 0.
Removing this 0 leaves a permitted string of leng... |
nemotron-math-v2-medium-0002488-fbd40c15-36bc-bc22-e79e-4b2355b5d52f | nvidia/Nemotron-Math-v2/aops | How many sets of natural numbers \(a, b, c, d\) exist such that \(a + b + c + d = 2015\)? | [
"We need number of solutions in natural numbers (positive integers) a,b,c,d such that sum = 2015. Usually natural numbers meaning positive integers (1,2,...). If includes zero? Usually natural numbers are positive but sometimes includes zero. Need interpret. They ask \"natural numbers\" likely positive integers (≥1... | To count the quadruples \((a,b,c,d)\) of natural numbers (i.e., positive integers) satisfying
\[
a+b+c+d = 2015,
\]
we use the “stars‑and‑bars’’ method.
---
### Stars and bars
For \(k\) positive integers whose sum is \(n\), the number of solutions is
\[
\binom{n-1}{k-1}.
\]
Here \(k=4\) and \(n=2015\). Hence... |
nemotron-math-v2-medium-0002502-e28480b5-5155-c15e-eab8-450743cdd19c | nvidia/Nemotron-Math-v2/aops | In how many ways can twelve pairs of twin brothers stand in a circle such that all pairs of twin brothers are neighbors? | [
"We have 12 pairs of twin brothers (total 24 individuals). They are to stand in a circle such that each twin pair stands adjacent (neighbors). So each pair must occupy two adjacent positions. The circle arrangement constraints: treat each pair as a block of two people, with internal order (which twin is left/right)... | Think of each pair of twins as a single “block’’ that must stay together.
1. **Arrange the blocks around the circle.**
There are 12 distinct blocks. Arrangements of \(n\) distinct objects in a circle are counted up to rotation, giving \((n-1)!\) possibilities.
\[
\text{Ways to place the blocks}= (12-1)! ... |
nemotron-math-v2-medium-0002505-28444cc2-a82b-1819-8c24-69b0e5934d6b | nvidia/Nemotron-Math-v2/aops | Solve the equation $a^2 + 10 = b^2$ in $\mathbb{Q}$. | [
"We need rational solutions (a,b) to a^2 + 10 = b^2. Equivalent to b^2 - a^2 = 10 => (b - a)(b + a) = 10. Over rationals, we need rational numbers x = b - a, y = b + a such that xy = 10, and then solving for a,b: b = (x + y)/2, a = (y - x)/2. So need rational x,y with product 10. Let x = p, y = 10/p for any nonzero... | \[
a^{2}+10=b^{2}\qquad (a,b\in\mathbb{Q})
\]
---
### 1. Factorisation
\[
b^{2}-a^{2}=10\quad\Longrightarrow\quad (b-a)(b+a)=10 .
\]
Denote
\[
x=b-a,\qquad y=b+a .
\]
Then \(x,y\in\mathbb Q\) and
\[
xy=10,\qquad x\neq0,\;y\neq0 .
\]
---
### 2. Parametrising the product \(xy=10\)
For any non‑zero ratio... |
nemotron-math-v2-medium-0002614-45f674f8-d055-6a67-8b88-71eee7b68010 | nvidia/Nemotron-Math-v2/aops | Evaluate the sum \(\sum_{k=2, k \text{ even}}^{\infty} \frac{1}{k!}\). | [
"We need sum over even k starting at 2: sum_{k even, k>=2} 1/k!. That's equal to sum_{n=1}^{∞} 1/( (2n)! ), because even numbers: k=2n with n>=1. So S = Σ_{n=1}∞ 1/(2n)!.",
"We know e^x = Σ_{m=0}∞ x^m/m!. For x=1, e = Σ_{m=0}∞ 1/m!. Separate into even and odd terms:",
"e = Σ_{n=0}∞ 1/(2n)! + Σ_{n=0}∞ 1/(2n+1)!.... | The even terms of the exponential series are
\[
\sum_{k=2,\,k\text{ even}}^{\infty}\frac1{k!}
=\sum_{n=1}^{\infty}\frac1{(2n)!}.
\]
Recall the Taylor series for the hyperbolic cosine,
\[
\cosh x =\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}.
\]
Setting \(x=1\),
\[
\cosh 1 = \sum_{n=0}^{\infty}\frac1{(2n)!}
... |
nemotron-math-v2-medium-0002951-774826b8-ab15-9b90-ed3f-66ca5fe016fe | nvidia/Nemotron-Math-v2/aops | Divide 96 into a sum of mutually coprime numbers using the smallest number of terms. | [
"We need to divide 96 into a sum of mutually coprime numbers using the smallest number of terms. I think it's partition 96 = a1 + a2 + ... + ak where each pair is coprime (gcd(ai, aj)=1 for all i≠j). Minimize k (number of terms). Find such a representation with minimal k.",
"We need to find smallest possible numb... | To minimize the number of summands we first ask whether the sum can be written with only **two** terms that are pairwise coprime.
Let
\[
96 = a + b ,\qquad a,b\in\mathbb Z_{>0},
\]
and require
\[
\gcd(a,b)=1 .
\]
Because \(96\) is even, both \(a\) and \(b\) must be odd (odd + odd = even).
Any two odd numbers... |
nemotron-math-v2-medium-0003420-a16eeac5-e22f-1e87-dac1-6da71cb1e88f | nvidia/Nemotron-Math-v2/aops | Find all functions \( f : \mathbb{R} \rightarrow \mathbb{R} \) that satisfy \( f(x+1) = 2f(x) \) for all real numbers \( x \). | [
"We need all functions f: R → R satisfying functional equation f(x+1) = 2 f(x) for all real x. This is like Cauchy-type periodic scaling, solution is f(x) = 2^x g({x}), where g is a periodic function of period 1. Indeed consider define h(x)=f(x)/2^x. Then h(x+1) = f(x+1)/2^{x+1} = 2 f(x) / 2^{x+1} = f(x)/2^x = h(x)... | Let
\[
h(x)=\frac{f(x)}{2^{x}}\qquad (x\in\mathbb {R}).
\]
Then
\[
h(x+1)=\frac{f(x+1)}{2^{x+1}}
=\frac{2f(x)}{2^{x+1}}
=\frac{f(x)}{2^{x}}
=h(x).
\]
Hence \(h\) is **1‑periodic**: \(h(x+1)=h(x)\) for every real \(x\).
Conversely, if \(h\) is any 1‑periodic function, define
\[
f(x)=2... |
nemotron-math-v2-medium-0003536-4f289e8f-7c41-7d52-50e7-deece9a19b6f | nvidia/Nemotron-Math-v2/aops | Find irrational numbers \( X \) and \( Y \) such that \( X^2 - Y \) is a natural number. | [
"We need to find irrational numbers X, Y such that X^2 - Y ∈ ℕ (natural numbers). So we need X^2 - Y = integer. Choose any irrational X and define Y = X^2 - n where n is a natural number. Then Y must be irrational as well. Need both irrational. But Y = X^2 - n; if X is irrational, X^2 could be rational? Possibly bu... | We need two irrational numbers \(X\) and \(Y\) such that
\[
X^{2}-Y\in \mathbb N .
\]
A convenient way to guarantee this is:
1. **Pick an irrational number \(X\) whose square is also irrational.**
This prevents \(X^{2}\) from becoming a rational integer after we subtract a natural number.
2. **Choose a natur... |
nemotron-math-v2-medium-0003751-0d03f80c-e50c-9d13-ceb9-39a0cf4c85f8 | nvidia/Nemotron-Math-v2/aops | Find the greatest possible value of the summands when any natural number \( n \) (where \( n \ge 3 \)) is expressed as a sum of several summands (not necessarily different). | [
"We need to parse: \"Find the greatest possible value of the summands when any natural number n (where n ≥ 3) is expressed as a sum of several summands (not necessarily different).\" Likely they ask: given any natural n ≥ 3, we express n as sum of several (>=2?) summands (maybe any number of summands). They ask \"g... | When we write a natural number \(n\;(n\ge 3)\) as a sum of several (at least two) natural numbers, we can make one of the summands as large as possible by making the other summands as small as possible.
* The smallest natural number is \(1\).
* If we use \(k\) summands, with the smallest possible values for the other ... |
nemotron-math-v2-medium-0003758-e99f3919-c104-9fc5-1487-caf463e1efc5 | nvidia/Nemotron-Math-v2/aops | Let \( a \) and \( b \) be real numbers. Find the maximum and minimum value of \( a \cos \phi + b \sin \phi \). | [
"We need to find the range of linear combination of sin and cos: a cos φ + b sin φ = R cos(φ - δ) or R sin(φ + ...). The max is sqrt(a^2 + b^2), min is -sqrt(a^2 + b^2). Provide reasoning.",
"Step-by-step: Use Cauchy-Schwarz or rewrite as R cos(φ - α). Compute R = sqrt(a^2+b^2). So maximum = R, minimum = -R.",
... | We want the extreme values of
\[
f(\phi)=a\cos\phi+b\sin\phi ,\qquad \phi\in\mathbb R,
\]
where \(a,b\) are fixed real numbers.
---
### 1. Rewrite as a single cosine
Write
\[
a\cos\phi+b\sin\phi=R\cos(\phi-\alpha)
\]
for suitable \(R>0\) and angle \(\alpha\).
Using the cosine addition formula
\[
R\cos(\... |
nemotron-math-v2-medium-0004302-2fc3d20f-144d-685b-8f98-31442fb1af5b | nvidia/Nemotron-Math-v2/aops | Write the function \( f(x) = x^x \) for \( x > 0 \) as a composition of two other functions. | [
"We need to express f(x)=x^x as composition of two functions: i.e., f = g ∘ h perhaps. Need to find g and h such that f(x) = g(h(x)). Common approach: use exponential and logarithm: x^x = e^{x \\ln x}. So define h(x) = x \\ln x, and g(u) = e^u. Then f(x) = g(h(x)). Provide reasoning.",
"We need to \"write the fun... | For \(x>0\) we can write
\[
x^{x}=e^{\;x\ln x}.
\]
Hence the function \(f(x)=x^{x}\) can be seen as the composition of two simpler functions:
\[
\begin{aligned}
h:&\;(0,\infty)\to\mathbb{R},\qquad h(x)=x\ln x,\\[4pt]
g:&\;\mathbb{R}\to(0,\infty),\qquad g(u)=e^{u}.
\end{aligned}
\]
Then
\[
f(x)=g\bigl(h(x)\bigr... |
nemotron-math-v2-medium-0004502-65185353-9f3c-51e2-b680-613efb095705 | nvidia/Nemotron-Math-v2/aops | For the polynomial \( f(x) = x^3 + px^2 + qx + r \) with roots \( a + bi \), \( a - bi \), and \( c \), compare the slopes of the tangent lines at the points \( (a, f(a)) \) and \( (c, f(c)) \). | [
"We need to compare slopes of tangent lines at x = a and x = c for cubic with complex conjugate roots a±bi and real root c. The polynomial f(x)=x^3+ p x^2 + q x + r has those roots, so factorization: (x-(a+bi))(x-(a-bi))(x-c)= ((x-a)^2 + b^2)(x-c). Expand to find derivative f'(x). Then compute slopes f'(a) and f'(c... | The cubic can be written from its roots
\[
f(x)=(x-(a+bi))(x-(a-bi))(x-c)=\bigl[(x-a)^2+b^{2}\bigr]\,(x-c).
\]
Let
\[
u(x)= (x-a)^2+b^{2},\qquad v(x)=x-c,
\]
so that \(f(x)=u(x)v(x)\).
Differentiate using the product rule:
\[
\begin{aligned}
f'(x) &= u'(x)v(x)+u(x)v'(x) \\
&= 2(x-a)(x-c)+\bigl[(x-a)^2+... |
nemotron-math-v2-medium-0004735-f959ccef-5006-80c8-047e-478b6466362c | nvidia/Nemotron-Math-v2/aops | Find two integers such that their product is $1$ less than three times their sum. | [
"We need integers x and y such that product = 1 less than three times their sum.",
"Equation: xy = 3(x + y) - 1. Rearr: xy - 3x - 3y + 1 = 0. Add 9 both sides? Actually rewrite as (x - 3)(y - 3) = 8? Let's check: Expand (x - 3)(y - 3) = xy -3x -3y +9. Compare with xy -3x -3y +1 =0 => xy -3x -3y +9 = 8. So (x-3)(y... | We are looking for integers \(x\) and \(y\) that satisfy
\[
xy = 3(x+y)-1 .
\]
---
### 1. Rearrange the equation
\[
xy-3x-3y+1=0 .
\]
Add \(9\) to both sides so that the left side factors:
\[
xy-3x-3y+9 = 8,
\]
\[
(x-3)(y-3)=8 .
\]
Thus the numbers \(x-3\) and \(y-3\) are two integers whose product equals \... |
nemotron-math-v2-medium-0004928-5949cf99-73c0-5808-30d2-3efcc27d4734 | nvidia/Nemotron-Math-v2/aops | Find all positive integers \( k \) for which the equation \( n^m - m^n = k \) has a solution in positive integers. | [
"We need all positive integers k such that there exist positive integers n,m with n^m - m^n = k. So the set of possible differences of exponential forms.",
"We need to characterize which k are achievable. Since n,m positive integers. Could be any integer? Let's explore.",
"Observation: If n=m then difference 0,... | Let
\[
n^{m}-m^{n}=k ,\qquad n,m\in\mathbb Z_{>0}.
\]
We have to determine for which positive integers \(k\) such a pair \((n,m)\) exists.
--------------------------------------------------------------------
### 1. Use the simplest possible exponent
Take the exponent \(m=1\).
Then
\[
n^{1}-1^{n}=n-1 .
\]
S... |
nemotron-math-v2-medium-0004929-dc720b39-d81b-d2ca-51a9-968e1dcb5638 | nvidia/Nemotron-Math-v2/aops | Let \( t_n \) denote the number of binary sequences of length \( n \) that do not contain three adjacent 1s. Determine a recurrence relation for \( t_n \), and provide enough initial values to generate the sequence. | [
"We need to derive recurrence for binary strings length n with no three consecutive 1's. This is classic linear recurrence: t_n = t_{n-1} + t_{n-2} + t_{n-3}. Actually need verify. Let's think.",
"Define strings avoiding \"111\". Count t_n.",
"We can classify strings by what they end with.",
"Let a_n be numbe... | Let
\[
t_n=\#\{\text{binary strings of length }n\text{ that contain no three consecutive }1\}.
\]
---
### Step 1 – Classify a string by its last symbols
A valid string of length \(n\) cannot end with “…111”.
Therefore its last block must be one of the following three possibilities
* it ends with a single \... |
nemotron-math-v2-medium-0005024-196cda54-9124-b8e7-0e81-763222c8bca1 | nvidia/Nemotron-Math-v2/aops | Find all continuous functions \( f(x) \) on \( \mathbb{R} \) such that \( f(x) = f(x+3) \). | [
"We need to find all continuous functions on real line that are periodic with period 3: f(x) = f(x+3). The condition is just periodic with period 3 (or perhaps period dividing 3). All continuous periodic functions with period 3, i.e., any function defined on [0,3] continuous, extended periodically. So answer: any c... | Because the functional equation
\[
f(x)=f(x+3)\qquad\forall x\in\mathbb R
\]
states that the value of \(f\) repeats every distance \(3\), any function that is **\(3\)-periodic** and continuous on \(\mathbb R\) satisfies the equation, and conversely any continuous solution must be \(3\)-periodic.
---
### Construct... |
nemotron-math-v2-medium-0005110-c5b1b644-0fc4-8729-0109-2441596b895d | nvidia/Nemotron-Math-v2/aops | Find the sum to infinity of the series \(\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \cdots\). | [
"We need sum_{k=1 to ∞} 1/( (2k)! ). Starting from k=1? First term 1/2! corresponds to k=1, yes. So sum_{k=1}∞ 1/(2k)!.",
"We can use expansion of cosh or cos. Recall e^x = sum_{n=0}∞ x^n / n!. Cosine: cos x = sum_{n=0}∞ (-1)^n x^{2n} / (2n)!. Hyperbolic cosine: cosh x = sum_{n=0}∞ x^{2n} / (2n)!. So sum over all... | The series is
\[
S=\frac1{2!}+\frac1{4!}+\frac1{6!}+\cdots=\sum_{k=1}^{\infty}\frac{1}{(2k)!}.
\]
Recall the Taylor series of the hyperbolic cosine:
\[
\cosh x = \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}.
\]
Setting \(x=1\),
\[
\cosh 1 = \sum_{n=0}^{\infty}\frac{1}{(2n)!}= \frac{e+e^{-1}}{2}.
\]
The term \(n=0\) ... |
nemotron-math-v2-medium-0005190-7570f91f-0aeb-fa44-8515-06b3f8089bd5 | nvidia/Nemotron-Math-v2/aops | Solve the equation $a^2 + 10 = b^2$ in $\mathbb{Q}$. | [
"We need rational solutions to a^2 + 10 = b^2 => b^2 - a^2 = 10 => (b-a)(b+a)=10. Over rationals, we can param. Let x = b-a, y = b+a, then xy=10, with x,y rational, and a = (y - x)/2, b = (x + y)/2. So any rational factorization of 10 yields solution. Since rational numbers can be expressed as x = 10/t, y = t for a... | \[
a^2+10=b^2\qquad (a,b\in\mathbb{Q})
\]
Rewrite the equation as a difference of squares
\[
b^2-a^2=(b-a)(b+a)=10 .
\]
Set
\[
x=b-a,\qquad y=b+a .
\]
Then \(x,y\in\mathbb{Q}\) and
\[
xy=10 .
\]
Any non‑zero rational number \(t\) gives a factorisation of \(10\) as
\[
x=\frac{10}{t},\qquad y=t .
\]
From... |
nemotron-math-v2-medium-0005633-b08d522d-9818-f4c9-4ee3-7a3bd01d4f86 | nvidia/Nemotron-Math-v2/aops | Divide 96 into a sum of mutually coprime numbers using the smallest number of terms. | [
"We need to divide 96 into a sum of mutually coprime numbers (pairwise coprime), using the smallest number of terms. We need to find a set of positive integers whose sum is 96, pairwise coprime, and the number of terms is minimized.",
"We must find minimum cardinality of a set of pairwise coprime positive integer... | To use the fewest possible summands we first ask whether a single summand could work.
A “division into a sum’’ requires at least two parts, so one term is not allowed.
Hence we look for a representation with **two** positive integers
\(a\) and \(b\) such that
* \(a+b = 96\)
* \(\gcd(a,b)=1\) (they are mutual... |
nemotron-math-v2-medium-0005677-fb0cd574-9c1a-1c9c-ad82-87392f118a8a | nvidia/Nemotron-Math-v2/aops | In how many ways may 5 people be seated in a row of twenty chairs given that no two people may sit next to one another? | [
"We need number of ways to choose 5 chairs out of 20 such that no two chosen chairs are adjacent, then assign 5 distinct people to those chairs (order matters). So count = (number of ways to select 5 nonadjacent positions) * 5!",
"Number of ways to choose k =5 positions from n=20 with no two consecutive: known as... | To seat the 5 distinct people we must first choose 5 chairs among the 20 so that no two chosen chairs are consecutive.
After the chairs are selected, the 5 people can be arranged in those chairs in \(5!\) ways.
---
### 1. Counting admissible sets of chairs
Let the chosen chairs have positions
\(1\le i_1<i_2<\... |
nemotron-math-v2-medium-0005678-cb8453cc-e869-ce5e-5238-cbfb57ccc724 | nvidia/Nemotron-Math-v2/aops | Given non-negative real numbers \(a, b, c, d, e, f, g\), find the maximum value of \(\max(a+b+c, b+c+d, c+d+e, e+f+g)\). | [
"We need to maximize the value of the maximum among four sums of three consecutive variables (but note the fourth is e+f+g, not d+e+f). Variables a,b,c,d,e,f,g are non-negative reals. No other constraints? Not given. So they can be arbitrarily large, leading to infinite maximum. Perhaps there is a hidden constraint... | Because the seven numbers are only required to be non‑negative, we can make any one of the four three‑term sums as large as we wish while keeping the other terms zero.
For example, let
\[
a = t,\qquad b=c=d=e=f=g = 0 ,
\]
with \(t>0\). Then
\[
\max (a+b+c,\; b+c+d,\; c+d+e,\; e+f+g)= a+b+c = t .
\]
Since \(t\)... |
nemotron-math-v2-medium-0005764-ea07677b-bbe8-955d-c117-915aa3b48ccf | nvidia/Nemotron-Math-v2/aops | In how many ways can the product of n distinct letters be formed? | [
"The question: \"In how many ways can the product of n distinct letters be formed?\" Likely means number of distinct words (arrangements) formed by concatenating n distinct letters? Or product meaning multiplication? It may be about forming a product like a1*a2*...*an where order matters? Could be permutations of n... | When we multiply \(n\) distinct letters (say \(a_{1},a_{2},\dots ,a_{n}\)) we may insert parentheses in many different ways.
Two parenthesizations are considered different even though multiplication is associative; for example
\[
(a_{1}a_{2})(a_{3}a_{4})\neq a_{1}(a_{2}a_{3}a_{4})
\]
The problem is therefore equi... |
nemotron-math-v2-medium-0005773-68537885-4ff6-1871-b927-7c869a8aa506 | nvidia/Nemotron-Math-v2/aops | A fair 12-sided die with sides 0 through 11 is rolled 4 times to form a base 12 number. Find \(p + q\) where \(\frac{p}{q}\) is the probability that \(11 \cdot 9 \cdot 8\) (base 10) divides the resulting number, and \(p\) and \(q\) are relatively prime positive integers. | [
"We have a 12-sided die with sides labeled 0-11 (base 12 digits). Roll 4 times to get a 4-digit base-12 number (including leading zeros? Likely yes). That forms a number N in base 12. Need probability that 11*9*8 (base 10) divides N. Compute product: 11*9=99, *8=792. So we need N divisible by 792.",
"We need prob... | The four rolls give four base‑12 digits
\[
N = a\cdot 12^{3}+b\cdot 12^{2}+c\cdot 12+d ,\qquad a,b,c,d\in\{0,1,\dots ,11\}.
\]
Every ordered 4‑tuple occurs with equal probability, so \(N\) is uniformly distributed among
\[
0,\;1,\;2,\;\dots ,\;12^{4}-1 = 20735 .
\]
The problem asks for the probability that
\... |
nemotron-math-v2-medium-0005814-2955ba10-9827-b06a-994b-d5b84c76dced | nvidia/Nemotron-Math-v2/aops | Given the set \( A = \{1, 2, 3, \ldots, 101\} \), determine the values of \( x_i \) such that the sets \( \{a + x_i \mid a \in A\} \) are considered. Provide two or three examples of possible sets \( \{a + x_i \mid a \in A\} \). | [
"The user asks: \"Given the set A = {1,2,3,...,101}, determine the values of x_i such that the sets {a + x_i | a ∈ A} are considered. Provide two or three examples of possible sets {a + x_i | a ∈ A}. Please reason step by step, and put your final answer within \\boxed{}.\"",
"Interpretation: For each shift x_i, w... | **Step‑by‑step reasoning**
1. The original set is
\[
A=\{1,2,3,\ldots ,101\}.
\]
2. For a given number \(x_i\) we form the “shifted’’ set
\[
A+x_i=\{\,a+x_i\mid a\in A\,\}.
\]
3. The value of \(x_i\) can be any integer (or, more generally, any real number) – the definition does not restrict it.
*If \(x_i\... |
nemotron-math-v2-medium-0006415-a2dd272c-6d39-295d-6726-5fc20f99d1bf | nvidia/Nemotron-Math-v2/aops | If \(\arctan a + \arctan b = 45^\circ\), then \(a + b =\) | [
"We need to solve given arctan a + arctan b = 45° (π/4). Find a + b? Probably expression in terms of a and b? Actually need numeric? There is relation: arctan a + arctan b = arctan((a+b)/(1-ab)) if sum not causing π/2 shift. So we set arctan((a+b)/(1-ab)) = 45° => (a+b)/(1-ab) = tan45° =1. So a+b = 1-ab. But then t... | We use the tangent–addition formula.
\[
\arctan a+\arctan b = \arctan\!\left(\frac{a+b}{1-ab}\right)
\]
provided the sum does not cross a vertical asymptote (which is the case here since the sum equals \(45^\circ\), a value where \(\tan\) is defined).
Given
\[
\arctan a + \arctan b = 45^\circ = \frac{\pi}{4},
\... |
nemotron-math-v2-medium-0006499-cec8ee53-7f6f-362e-7091-925d6bf5c315 | nvidia/Nemotron-Math-v2/aops | Find the set of points \( p(x, y) \) in the complex plane such that \( \left| \sqrt{x^2 + 4} + i \sqrt{y - 4} \right| = 10 \). | [
"We need to find the set of points (x,y) in the complex plane such that the magnitude of complex number sqrt(x^2 + 4) + i sqrt(y - 4) equals 10.",
"Let’s denote A = sqrt(x^2 + 4), B = sqrt(y - 4). Then the complex number is A + i B. The magnitude is sqrt(A^2 + B^2) = sqrt( (x^2+4) + (y-4) )? Wait, A^2 = x^2+4. B^... | Let
\[
z=\sqrt{x^{2}+4}+i\sqrt{y-4},
\qquad
A=\sqrt{x^{2}+4},\; B=\sqrt{y-4}\;( \ge 0).
\]
The modulus of \(z\) is
\[
|z|=\sqrt{A^{2}+B^{2}}
=\sqrt{(x^{2}+4)+(y-4)}
=\sqrt{x^{2}+y}.
\]
The condition \(\;|z|=10\;\) therefore gives
\[
\sqrt{x^{2}+y}=10 \Longrightarrow x^{2}+y=100.
\]
The square‑r... |
nemotron-math-v2-medium-0006558-ceea41ea-2953-ddfc-a9ee-5c3d71a9c7a7 | nvidia/Nemotron-Math-v2/aops | In a circle with $n$ points, each point can be painted either black or white. Calculate the number of distinct ways to paint the circle, considering that two colorings are considered the same if one can be obtained from the other by rotation. | [
"We need to count number of binary colorings of n labeled points on a circle modulo rotation symmetry (cyclic group C_n). So count of orbits under rotation action. Use Burnside's Lemma (Polya's enumeration). Number of colorings fixed by rotation by k positions (where k=0,...,n-1). A rotation by k (with gcd(k,n)=d) ... | Let the \(n\) points be labelled \(0,1,\dots ,n-1\) around the circle.
A colouring is a function
\[
c:\{0,1,\dots ,n-1\}\longrightarrow\{\,\text{black},\text{white}\,\},
\]
so there are \(2^{n}\) colourings in total.
Two colourings are considered the same if one can be obtained from the other by a rotation.
T... |
nemotron-math-v2-medium-0006919-7903f67d-1fd3-2e12-db45-8227f5688d8e | nvidia/Nemotron-Math-v2/aops | Create a Diophantine equation that has $(p,q) = (3,7)$ as its solution, where $p$ and $q$ are natural numbers (or prime numbers). | [
"We need a Diophantine equation that has (p,q)=(3,7) as its solution. Likely they want an equation in p and q with natural (or prime) solutions, and (3,7) is a solution. Could be many. Perhaps they want a specific equation that uniquely gives that solution. For example, use linear combination: 2p + q = 13 gives 2*3... | **Step‑by‑step construction**
1. **Goal.**
We want an equation in the two unknowns \(p\) and \(q\) whose only solution in the set of natural (or prime) numbers is
\[
(p,q)=(3,7).
\]
2. **Using squares to force equality.**
For any integer \(x\), the square \(x^{2}\) is non‑negative, and the sum of... |
nemotron-math-v2-medium-0007327-dfc9e0ec-c56b-20c0-a9d9-f426d2bd2aeb | nvidia/Nemotron-Math-v2/aops | Using sigma notation, find the number and the sum of its digits for a number in base \( q \) represented by \( a_{n}a_{n-1}a_{n-2}...a_{3}a_{2}a_{1}a_{0} \). | [
"We need to interpret the question: \"Using sigma notation, find the number and the sum of its digits for a number in base q represented by a_n a_{n-1}... a_0\". They likely want expression for the value of the number (its numeric value) and sum of its digits (i.e., sum_{i=0}^{n} a_i). The \"number\" maybe the inte... | Let
\[
N = a_{n}a_{n-1}a_{n-2}\dots a_{3}a_{2}a_{1}a_{0}
\qquad (a_i\in\{0,1,\dots ,q-1\})
\]
be a number written in base \(q\) (the left‑most digit \(a_n\neq 0\)).
When we expand the positional notation we obtain the ordinary (base‑10) value of the number:
\[
\boxed{N \;=\; \sum_{i=0}^{n} a_i\,q^{\,i}}
\]
Inde... |
nemotron-math-v2-medium-0007361-f708b9d8-b1ae-0315-ff36-673c43cb1c85 | nvidia/Nemotron-Math-v2/aops | There are $n$ people standing in $m$ stations, forming $m$ queues. Two schemes are considered different if there exists a station whose queue consists of different people or has a different order. How many different schemes are there? | [
"We need to interpret the problem: There are n people standing in m stations, forming m queues. So we have n distinct people, and we distribute them among m stations (queues). Each station gets a possibly empty queue; the order within each queue matters. So essentially we are partitioning an ordered sequence (permu... | We have \(n\) distinct people and \(m\) distinct stations (queues).
For a scheme we must decide
* which station each person stands in, and
* the order of the people inside each station.
---
### Step 1 – Order all people first
Take any permutation of the \(n\) people.
There are
\[
n!
\]
such permutatio... |
nemotron-math-v2-medium-0007795-67434740-21dd-be62-0855-196c30ea31c9 | nvidia/Nemotron-Math-v2/aops | Two of the numbers in a three-term combination are 17 and 24. There are 40 possibilities for the third number. If the order of the numbers is unknown, how long will it take to try all possible combinations at a rate of ten seconds per try? | [
"We have a three-term combination lock. Two numbers are known: 17 and 24. The third number could be any of 40 possibilities (presumably 40 distinct numbers). The order of numbers is unknown: the combination could be any permutation of the three numbers. Need to try all possible combos at 10 seconds per try. Compute... | We know two of the three numbers: 17 and 24.
The third number could be any one of 40 possibilities.
Since we don’t know the order of the three numbers, for each possible third number we must try every permutation of the three distinct numbers.
With three distinct numbers there are
\[
3! = 6
\]
possible orders.... |
nemotron-math-v2-medium-0008075-49cf1036-e209-50c8-514b-4ecb87a4f471 | nvidia/Nemotron-Math-v2/aops | Find the prime number \( p \) such that \( p^2 \) divides \( (p-1)! + 1 \). | [
"We need prime p such that p^2 divides (p-1)! + 1. This is reminiscent of Wilson's theorem: (p-1)! ≡ -1 mod p, so (p-1)! + 1 ≡ 0 mod p. So p divides (p-1)!+1 always for prime p. The question asks for p such that p^2 also divides (p-1)!+1, i.e., (p-1)! ≡ -1 mod p^2. Those primes are called Wilson primes. Known Wilso... | Recall **Wilson’s theorem**:
\[
(p-1)!\equiv -1\pmod p\qquad\text{for every prime }p .
\]
Hence
\[
(p-1)!+1\equiv 0\pmod p ,
\]
so the expression is always divisible by \(p\).
The problem asks for a stronger condition:
\[
p^{2}\mid (p-1)!+1\qquad\Longleftrightarrow\qquad (p-1)!\equiv -1\pmod{p^{2}} .
\]
A pri... |
nemotron-math-v2-medium-0008277-c1befeea-b758-0d7e-6542-10f183b3372d | nvidia/Nemotron-Math-v2/aops | Find a recurrence relation for the Bell number $b_n$, which represents the number of ways to partition the set $\{1,2,\ldots,n\}$. | [
"We need to give recurrence for Bell numbers. Known recurrence: b_{n+1} = sum_{k=0}^{n} \\binom{n}{k} b_k. Also exponential generating function: B(x)=exp(e^x-1). Also Dobinski's formula. They ask for recurrence relation; classic recurrence: b_{n+1} = \\sum_{k=0}^{n} \\binom{n}{k} b_k. Also can derive from adding el... | The Bell numbers count the set partitions of
\[
[n]=\{1,2,\dots ,n\}.
\]
Let \(b_n\) denote the number of partitions of \([n]\) (with the convention \(b_0=1\) for the empty set).
---
### Adding a new element
Consider the set \([n+1]=\{1,2,\dots ,n,n+1\}\).
In a partition of \([n+1]\) the element \(n+1\) must... |
nemotron-math-v2-medium-0008420-f7540079-5fa6-e556-e728-b81cf1eb3799 | nvidia/Nemotron-Math-v2/aops | An ordinary deck of 52 cards with 4 aces is shuffled, and then the cards are drawn one by one until the first ace appears. On average, how many cards are drawn? | [
"We need expected position of first ace in random permutation of 52 containing 4 aces. This is classic: expectation of minimum of 4 random positions (distinct) uniformly from 1..52. Equivalent to: choose 4 distinct integers uniformly; order statistics. Expected minimum = (n+1)/(k+1) where n=52, k=4? Actually for or... | Let
\[
X=\text{position of the first ace in a random permutation of the 52 cards}
\]
(so \(X\) is the number of cards we draw, including the ace).
---
### 1. Distribution of \(X\)
The event \(\{X>m\}\) means that the first \(m\) cards contain **no** ace.
We have 48 non‑aces and 4 aces. The number of ways to c... |
nemotron-math-v2-medium-0008486-89e8f7a7-f06c-414e-dde1-79ea3757b1d6 | nvidia/Nemotron-Math-v2/aops | Find all polynomials \( P(x) \) such that \( P(x) = P(1-x) \). | [
"We need all polynomials satisfying functional equation P(x)=P(1-x) for all x (real). This indicates symmetry about x=1/2. So any polynomial that is symmetric under x -> 1-x. Let y = x - 1/2. Then 1 - x = 1 - (y+1/2) = 1/2 - y. So condition becomes Q(y) = P(y+1/2) = P(1 - (y+1/2)) = P(1/2 - y) = Q(-y). So Q is an e... | Let
\[
Q(y)=P\!\left(y+\frac12\right),\qquad y\in\mathbb R .
\]
Then
\[
P(1-x)=P\!\left(1-\Big(y+\frac12\Big)\right)
=P\!\left(\frac12-y\right)=Q(-y).
\]
Hence the condition \(P(x)=P(1-x)\) becomes
\[
Q(y)=Q(-y)\qquad\text{for all }y .
\]
So \(Q\) is an **even** polynomial.
Every even polynomial can... |
nemotron-math-v2-medium-0008505-a164e25b-ae8c-3ca9-c25f-82a522aa4276 | nvidia/Nemotron-Math-v2/aops | Given the set \( A = \{1, 2, 3, \ldots, 101\} \), determine the values of \( x_i \) such that the sets \( \{a + x_i \mid a \in A\} \) are considered. Provide two or three examples of possible sets \( \{a + x_i \mid a \in A\} \). | [
"We need to parse: \"Given the set A={1,2,3,...,101}, determine the values of x_i such that the sets {a + x_i | a ∈ A} are considered. Provide two or three examples of possible sets {a + x_i | a ∈ A}.\"",
"Interpretation: They want to pick some integer shifts x_i and then consider the translated sets. Maybe ask t... | **Step‑by‑step reasoning**
1. **The original set**
\[
A=\{1,2,3,\ldots ,101\}.
\]
2. **What does “\(a+x_i\)” mean?**
For a fixed number \(x_i\) (the *shift* or *translation*), we add that same number to every element of \(A\).
The resulting set is
\[
A+x_i=\{a+x_i\mid a\in A\}.
\]
3. ... |
nemotron-math-v2-medium-0008517-ae5cac6d-d6ed-360c-a1f8-0ffdb6489b96 | nvidia/Nemotron-Math-v2/aops | Find all positive integers \( k \) for which the equation \( n^m - m^n = k \) has a solution in positive integers. | [
"We need to find all positive integers k such that the Diophantine equation n^m - m^n = k has a solution in positive integers n,m (both positive). So we need to characterize possible differences.",
"We look for all k>0 that can be expressed as difference of two exponential terms n^m and m^n. Note that n and m are... | We want all positive integers \(k\) for which
\[
n^{\,m}-m^{\,n}=k \qquad (n,m\in \mathbb{Z}_{>0})
\]
has a solution.
--------------------------------------------------------------------
### 1. Use the trivial exponent \(m=1\)
Take \(m=1\).
Then the equation becomes
\[
n^{1}-1^{\,n}=n-1=k .
\]
For any inte... |
nemotron-math-v2-medium-0008524-3921f201-d06d-3a32-217c-2ffd55688412 | nvidia/Nemotron-Math-v2/aops | Determine the minimal possible value $T$ for a square $(T \times T)$-seed resulting in a co-prime $(p, q)$-balanced array, when $p, q$ are both prime. | [
"We need to parse the problem. It asks: \"Determine the minimal possible value $T$ for a square $(T \\times T)$-seed resulting in a co-prime $(p, q)$-balanced array, when $p, q$ are both prime. Please reason step by step, and put your final answer within \\boxed{}.\"",
"We need to interpret what is a \"(p, q)-bal... | To obtain a \((p,q)\)-balanced array from a square seed we must be able to tile the seed with
\(p\times q\) blocks.
Because each block must fit exactly both horizontally and vertically, the side length
\(T\) of the seed has to be a common multiple of the two dimensions \(p\) and \(q\).
Since \(p\) and \(q\) are dis... |
nemotron-math-v2-medium-0008773-c027e2bc-abee-ed05-14b4-e1e4b7c52c36 | nvidia/Nemotron-Math-v2/aops | In how many ways can twelve pairs of twin brothers stand in a circle such that all pairs of twin brothers are neighbors? | [
"We have 12 pairs of twin brothers, total 24 individuals. They must stand in a circle such that each pair of twins are adjacent. Count number of arrangements (circular permutations) where each twin pair is adjacent.",
"Interpretation: Twins are distinct (brother A vs brother B), but they are considered as individ... | Let the twins be \((T_1^{(1)},T_1^{(2)}),\,(T_2^{(1)},T_2^{(2)}),\dots ,(T_{12}^{(1)},T_{12}^{(2)})\).
---
### 1. Treat each twin pair as a block
Since each pair must occupy two neighbouring seats, we first arrange the 12 blocks around the circle.
For a circular arrangement of \(12\) distinct blocks, rotations... |
nemotron-math-v2-medium-0008807-866e174f-b602-c43c-9b62-a8db34574ff6 | nvidia/Nemotron-Math-v2/aops | Determine the minimal possible value $T$ for a square $(T \times T)$-seed resulting in a co-prime $(p, q)$-balanced array, when $p, q$ are both prime. | [
"We need to parse the problem: \"Determine the minimal possible value $T$ for a square $(T \\times T)$-seed resulting in a co-prime $(p, q)$-balanced array, when $p, q$ are both prime.\"",
"We need context: square seed generating a co-prime (p,q)-balanced array? Likely about constructing arrays where p and q are ... | **Step‑by‑step reasoning**
1. A square \((T\times T)\)-seed is repeated (tiled) to generate a larger array.
2. For the resulting array to be \((p,q)\)-balanced, the pattern must repeat with a period that is simultaneously a multiple of \(p\) (in one direction) and of \(q\) (in the other direction).
3. Hence the si... |
nemotron-math-v2-medium-0009607-65a21a46-07db-890b-0a58-ca9da308e1c8 | nvidia/Nemotron-Math-v2/aops | Compute \( \lim_{n \to \infty} \sum_{i=1}^n S_{i} \), where \( S_{i} \) is the area of triangle \( A_{i}B_{i}C_{i} \) formed by the midpoints of the sides of triangle \( A_{i-1}B_{i-1}C_{i-1} \). | [
"We need to understand the problem. We have a sequence of triangles. Starting with some initial triangle A_0 B_0 C_0? Not specified. Probably given initial triangle. They define S_i as area of triangle formed by midpoints of sides of previous triangle. So each iteration forms medial triangle. The medial triangle ha... | The triangle formed by joining the mid‑points of the sides of any triangle is similar to the original triangle with similarity ratio \(\frac12\).
Hence its area is reduced by a factor \(\left(\frac12\right)^2 = \frac14\).
Let
\[
S_0 = \text{area of the original triangle } A_0B_0C_0 .
\]
Then
\[
S_1=\frac14 S_... |
nemotron-math-v2-medium-0010062-7409da83-89e3-272f-01e6-77b82b6aa6df | nvidia/Nemotron-Math-v2/aops | Solve the recurrence relation $a_n = -6a_{n-1} - 9a_{n-2}$ with initial conditions $a_0 = 3$ and $a_3 = 6$ by giving an explicit formula for $a_n$. | [
"We need to solve recurrence a_n = -6 a_{n-1} -9 a_{n-2}. This is linear homogeneous constant coefficients second order. Characteristic equation: r^2 = -6 r -9, i.e., r^2 + 6 r + 9 = 0 => (r+3)^2=0. So double root r = -3 multiplicity 2.",
"General solution: a_n = (C1 + C2 n) (-3)^n.",
"We have initial condition... | The characteristic equation of
\[
a_n = -6a_{n-1}-9a_{n-2}
\]
is
\[
r^{2}+6r+9=0\;\Longrightarrow\;(r+3)^2=0,
\]
so the root \(r=-3\) has multiplicity \(2\).
Hence the general term is
\[
a_n=(C_1+C_2 n)(-3)^n .
\]
Use the initial conditions.
* For \(n=0\): \(a_0=C_1=3\) ⇒ \(C_1=3\).
* For \(n=3\): \(a_3... |
nemotron-math-v2-medium-0010333-388acaaf-4ebd-1358-1ed0-e6c42f2f1ddf | nvidia/Nemotron-Math-v2/aops | Let \( f(x) \) be a 98-degree polynomial such that \( f(x) = \frac{1}{x} \) for \( x = 1, 2, 3, \ldots, 99 \). Find \( f(100) \). | [
"We need to find polynomial of degree 98 that matches 1/x at integer points 1..99. Define g(x)= x f(x) -1. Then g is degree 99? Let's see: f is degree 98, multiply by x gives degree 99. Subtract 1 gives degree at most 99. For x = 1..99, we have f(x)=1/x, so x f(x) -1 =0, thus g(x) has roots at 1,...,99. So g(x) is ... | Let
\[
g(x)=x\,f(x)-1 .
\]
Since \(f\) is a polynomial of degree \(98\), \(g\) is a polynomial of degree at most \(99\).
For each integer \(x=1,2,\dots ,99\),
\[
f(x)=\frac1{x}\quad\Longrightarrow\quad x f(x)-1=0,
\]
so \(g(x)=0\) at those 99 points. Hence \(g\) has the factorization
\[
g(x)=k\,(x-1)(x-2)\cdo... |
nemotron-math-v2-medium-0010365-5a50aeae-7d61-a01b-d6f7-feef81ecb744 | nvidia/Nemotron-Math-v2/aops | Find an expression with a range equal to the set $\{1,3\}$. | [
"The user asks: \"Find an expression with a range equal to the set {1,3}. Please reason step by step, and put your final answer within \\boxed{}.\"",
"Interpretation: They want a function (or expression) whose set of possible outputs (range) is exactly {1,3}. So we need an expression that yields either 1 or 3 for... | **Step 1 – What we need**
We want a formula (a function or expression) whose possible output values are **exactly** the two numbers \(1\) and \(3\).
Thus the range of the expression must be the set \(\{1,3\}\).
**Step 2 – Choose a simple periodic pattern**
A convenient way to obtain two alternating values is to ... |
nemotron-math-v2-medium-0010669-bf993e18-f351-4fa5-eb1a-7c396fe0f208 | nvidia/Nemotron-Math-v2/aops | Solve for $x$ in terms of $C$ in the equation $x + \sqrt{x} = C$. | [
"We need to solve x + sqrt(x) = C, for x in terms of C. Assume C is real, and x≥0 (since sqrt). Solve equation: let t = sqrt(x) => t ≥0, x = t^2. Then t^2 + t = C => t^2 + t - C = 0. Solve quadratic in t: t = [-1 ± sqrt(1 + 4C)] / 2. Since t = sqrt(x) ≥ 0, we need the non-negative root. The discriminant D = 1+4C. N... | Let
\[
t=\sqrt{x}\qquad (t\ge 0)
\]
Then \(x=t^{2}\) and the given equation becomes
\[
t^{2}+t=C\;\Longrightarrow\;t^{2}+t-C=0 .
\]
Solve this quadratic for \(t\):
\[
t=\frac{-1\pm\sqrt{1+4C}}{2}.
\]
Because \(t=\sqrt{x}\ge0\), we keep the “\(+\)” root:
\[
\sqrt{x}=t=\frac{-1+\sqrt{1+4C}}{2}.
\]
Now square ... |
nemotron-math-v2-medium-0010761-6b80c0aa-aedc-c477-ee65-be29dde7b8da | nvidia/Nemotron-Math-v2/aops | Annah has a serious and genuine honesty policy for the AOIME. If Annah sees a geometry problem on problem number $n$, the probability she uses geogebra for that problem is $\frac{n}{15}$. The AOIME has $14$ non-geo problems and #$15$ is a geo problem. Calculate the expected score for Annah on the AOIME, assuming that A... | [
"We need to parse the problem. Annah's honesty policy: If she sees a geometry problem on problem number n, the probability she uses geogebra for that problem is n/15. The AOIME has 14 non-geo problems and #15 is a geo problem. So there are 15 problems total, problem numbers 1-15. Only problem 15 is geometry. So for... | The AOIME consists of 15 problems numbered \(1,2,\dots ,15\).
- Only problem \(\#15\) is a geometry problem.
- For a geometry problem numbered \(n\), Annah uses GeoGebra with probability \(\frac{n}{15}\).
Hence for problem \(\#15\),
\[
P(\text{use GeoGebra})=\frac{15}{15}=1 .
\]
According to the policy, Annah gets ... |
nemotron-math-v2-medium-0010762-7470c576-1439-12d1-0b60-4cb85e699b82 | nvidia/Nemotron-Math-v2/aops | Given a polynomial \( P(x) \), a prime number \( q \), and natural numbers \( \alpha \) and \( n \), if \( P(2^n) = q^{\alpha} \), determine the form of the polynomial \( P(x) \). | [
"We need to determine the form of P(x) given that P evaluated at 2^n equals q^α. Likely the conclusion is that P(x)=q^α for all x? But constraints: polynomial with integer coefficients? Not specified. Probably they want that P(x) must be constant polynomial equal to q^α, because otherwise for varying n you'd get di... | Because we only know the value of the polynomial at the single point \(x=2^{n}\), the only restriction on \(P(x)\) is that it takes the prescribed value there.
Any polynomial can be written as a constant term plus a multiple of the factor \((x-2^{n})\). Hence
\[
P(x)=q^{\alpha}+ (x-2^{n})\;R(x),
\]
where \(R(x)\) i... |
nemotron-math-v2-medium-0011089-3784bfbb-50fb-f737-7102-0b4b2d6c0392 | nvidia/Nemotron-Math-v2/aops | Find the sign of the expression \(4^{16} + 3^{12} - 6^{4} - 24^{4} - 8\). | [
"We need sign of number: 4^16 + 3^12 - 6^4 - 24^4 - 8.",
"Compute or compare magnitudes.",
"First compute values:",
"4^16 = (2^2)^16 = 2^32. 2^10=1024, 2^20≈1,048,576, 2^30≈1,073,741,824. 2^32 = 4,294,967,296.",
"3^12 = (3^6)^2 = 729^2 = 531,441.",
"6^4 = 1296? Wait 6^3=216, 6^4=1296.",
"24^4: 24^2 = 57... | First rewrite each term with convenient bases.
\[
\begin{aligned}
4^{16} &= (2^{2})^{16}=2^{32}=4\,294\,967\,296,\\[2pt]
3^{12} &= (3^{6})^{2}=729^{2}=531\,441,\\[2pt]
6^{4} &= 1296,\\[2pt]
24^{4} &= (24^{2})^{2}=576^{2}=331\,776.
\end{aligned}
\]
Now combine the positive terms and the negative terms separately:
\[... |
nemotron-math-v2-medium-0011277-a2181296-8ce2-38d4-7cfa-f2ffe6c44132 | nvidia/Nemotron-Math-v2/aops | Annah has a serious and genuine honesty policy for the AOIME. If Annah sees a geometry problem on problem number $n$, the probability she uses geogebra for that problem is $\frac{n}{15}$. The AOIME has $14$ non-geo problems and #$15$ is a geo problem. Calculate the expected score for Annah on the AOIME, assuming that A... | [
"We need to parse: Annah has honesty policy: if she sees a geometry problem on problem number n, probability she uses geogebra for that problem is n/15. The AOIME has 14 non-geo problems and #15 is a geo problem. So total 15 problems, problem numbers 1-14 are non-geometry, problem 15 is geometry.",
"She will alwa... | **Step‑by‑step reasoning**
1. **Identify the problems:**
- The AOIME has 15 problems total.
- Problems 1‑14 are *non‑geometry* problems.
- Problem 15 is the *only* geometry problem.
2. **Probability of using GeoGebra:**
- Annah’s honesty policy only gives a probability for geometry problems:
... |
nemotron-math-v2-medium-0011429-a511b9a6-9bae-f413-f0e3-7027ee4a5ded | nvidia/Nemotron-Math-v2/aops | What is the number of surjective maps from a set \( E \) to a set \( F \) with cardinalities \( n \) and \( m \) respectively? | [
"We need to answer: number of surjective (onto) functions from an n-element set to an m-element set. Known formula: m! * S(n,m), where S(n,m) is Stirling numbers of the second kind (number of ways to partition n labeled elements into m nonempty unlabeled subsets), then assign those subsets to the m target elements ... | Let
* \(|E|=n\) and \(|F|=m\) (the elements of \(E\) are labelled, those of \(F\) are also labelled),
* a map \(f:E\to F\) is **surjective** (onto) iff every element of \(F\) is hit by at least one element of \(E\).
---
### 1. Count all functions
Without any restriction a function can send each of the \(n\) e... |
nemotron-math-v2-medium-0011803-f00cadff-60bd-b563-4ba3-322051f5e2f6 | nvidia/Nemotron-Math-v2/aops | Find the value of \( A \) if \( \tan 4A = \cot 6A \). | [
"We need solve tan 4A = cot 6A. Cot is 1/tan, but also cotθ = tan(π/2 - θ). So equation tan4A = cot6A => tan4A = tan(π/2 - 6A) + possibly plus kπ because tan periodic.",
"Thus: 4A = π/2 - 6A + kπ, for integer k.",
"Solve: 4A + 6A = π/2 + kπ => 10A = π/2 + kπ = (π/2)(1 + 2k). So A = (π/20)(1 + 2k) = (π/20) + (π/... | Since
\[
\cot 6A=\frac{1}{\tan 6A}= \tan\!\Bigl(\frac{\pi}{2}-6A\Bigr),
\]
the equation
\[
\tan 4A = \cot 6A
\]
becomes
\[
\tan 4A = \tan\!\Bigl(\frac{\pi}{2}-6A\Bigr).
\]
Two tangents are equal when their arguments differ by an integer multiple of \(\pi\):
\[
4A = \frac{\pi}{2} - 6A + k\pi,\qquad k\in\mat... |
nemotron-math-v2-medium-0011875-06167934-8916-1bae-8290-b2310b5dbb13 | nvidia/Nemotron-Math-v2/aops | Given an $m \times n$ chessboard and $k$ rooks, determine the number of ways to place the rooks on the chessboard such that no two rooks attack each other, where $k \le m$ and $k \le n$. | [
"We need to count placements of k non-attacking rooks on an m x n board, with k ≤ m, n. Standard combinatorial result: choose k rows from m, choose k columns from n, then assign a bijection between selected rows and columns (i.e., place one rook per selected row and column). Number = C(m, k) * C(n, k) * k! . Equiva... | **Step‑by‑step reasoning**
1. **Choose the rows that will contain rooks.**
Since at most one rook may sit in any row, we must pick exactly \(k\) distinct rows from the \(m\) rows of the board.
\[
\text{Number of ways} = \binom{m}{k}.
\]
2. **Choose the columns that will contain rooks.**
Similarly... |
nemotron-math-v2-medium-0011982-30788b78-f5df-354b-34ba-51da2b810ed2 | nvidia/Nemotron-Math-v2/aops | Find a pair of integers $(x, y)$ such that $15x^2 + y^2 = 2^{2000}$. | [
"We need integer solutions to 15 x^2 + y^2 = 2^{2000}. A huge power of 2; we need find any integer pair. Means we need to represent 2^{2000} as sum of a square and 15 times a square. Need to find x, y integers possibly with large magnitude. Use modular constraints.",
"Observation: equation resembles representatio... | \[
15x^{2}+y^{2}=2^{2000}
\]
Taking \(x=0\) gives
\[
y^{2}=2^{2000}\qquad\Longrightarrow\qquad y=\pm 2^{1000}.
\]
Hence a pair of integers satisfying the equation is
\[
\boxed{(x,y)=(0,\;2^{1000})}\qquad\text{(or }(0,-2^{1000})\text{).}
\] |
End of preview. Expand in Data Studio
Nemotron Math V2 Medium Mini
A compact textual reasoning dataset derived from the medium split of nvidia/Nemotron-Math-v2 at immutable revision 8e793210e175b6406c752a870f585f62de98c0d3.
Selection boundary
This extraction intentionally excludes tool-use semantics:
- Accept exactly two messages with roles
userthenassistant. - Reject records declaring tools, containing assistant tool calls, containing tool-result messages, or containing any additional turns.
- Require the injected source user message to end with the exact
problemtext. - Read reasoning only from the sole assistant message's
reasoning_content. - Read the final response only from that assistant message's
content. - Preserve pinned
medium-split row order without sampling or shuffling.
Each output row contains id, source, question, steps, and answer. Reasoning is split on blank lines into whitespace-stripped steps. Source UUIDs are not globally unique across trajectories, so every output ID combines the immutable zero-padded source row position with the source UUID.
Rows must contain 3-50 steps, at least 128 total reasoning characters, and no step longer than 1,024 characters. The exact Llama 3 training sequence must be at most 4,096 characters and 1,024 tokens under the prepared models/llama3-1.2b tokenizer.
Dataset statistics
| Metric | Value |
|---|---|
| Source rows scanned | 2,502,305 |
| Final records | 66,344 |
| Retention | 2.651% |
| File size (canonical JSONL) | 172,915,780 bytes |
| Sources represented | 2 |
| Unique questions | 25,275 |
| Total reasoning steps | 424,750 |
| Mean steps per record | 6.40 |
| Median steps per record | 6 |
| Steps per record | 3-45 |
| Mean individual step length | 167.40 characters |
| Median individual step length | 96 characters |
| Individual step-length range | 1-1,024 characters |
| Median total reasoning length | 1,011 characters |
| Total reasoning-length range | 165-3,561 characters |
| Answer-length range | 15-3,006 characters |
| Question-length range | 14-1,222 characters |
Source composition
| Source | Records |
|---|---|
nvidia/Nemotron-Math-v2/stackflow |
57,892 |
nvidia/Nemotron-Math-v2/aops |
8,452 |
| Total | 66,344 |
Rejection accounting
| Reason | Rows |
|---|---|
| Source conversation is not exactly two turns | 762,942 |
| Tool trajectory | 463,996 |
| Complete Llama 3 sequence over 4,096 characters | 404,838 |
| More than 50 reasoning steps | 338,489 |
| A reasoning step over 1,024 characters | 309,154 |
| Complete Llama 3 sequence over 1,024 tokens | 122,303 |
| Fewer than 3 reasoning steps | 34,031 |
| User message does not end with exact problem | 208 |
| Accepted | 66,344 |
| Total | 2,502,305 |
Integrity and provenance
The canonical local JSONL has SHA-256 9840f0263be78e9ee6648287ee1a758f7fd40ee72e6d1012aeae780011fabe4a. The repository includes construction_manifest.json with the pinned source revision, tool-exclusion boundary, ID scheme, complete filter contract, rejection counts, and content hash.
Rows retain the source dataset's row-level Creative Commons licensing. Usage remains subject to the source dataset's licensing terms and documented limitations.
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