Datasets:
id stringlengths 66 66 | source stringclasses 2
values | question stringlengths 14 1.1k | steps listlengths 3 46 | answer stringlengths 9 3.14k |
|---|---|---|---|---|
nemotron-math-v2-low-00000009-bf8daf9b-faa3-6b0c-a8ec-85a504a12694 | nvidia/Nemotron-Math-v2/aops | What is the total number of squares that can be drawn on an \( m \times n \) chessboard? | [
"We need formula: sum_{k=1}^{min(m,n)} (m - k +1)(n - k +1). Compute closed form? It's = sum_{k=1}^{p} (m - k +1)(n - k +1) where p=min(m,n). Expand: sum ( (m+1 - k)(n+1 - k) ) = sum ( (m+1)(n+1) - (m+1)k - (n+1)k + k^2 ). = p*(m+1)(n+1) - (m+1 + n+1) * sum k + sum k^2. Sum k = p(p+1)/2, sum k^2 = p(p+1)(2p+1)/6. S... | The number of axis‑parallel squares that can be drawn on an \(m\times n\) rectangular grid (with \(m\) rows and \(n\) columns) is obtained by counting, for each possible side‑length \(k\), how many positions that square can occupy.
---
### 1. Count squares of a fixed side‑length
If the square has side‑length \(k\... |
nemotron-math-v2-low-00000011-728e138a-ff1d-ee43-81ca-46f7b27cb157 | nvidia/Nemotron-Math-v2/aops | Let an integer quadrilateral be a quadrilateral whose sides are all of integer length, whose diagonals form $4$ segments of integer length and whose area is also an integer. Find the area of an integer quadrilateral with at least $3$ sides of length $13$. | [
"We need quadrilateral with sides integers, diagonals split into 4 integer segments (so both diagonals integer and their intersection yields segments? Means the two diagonals intersect, forming 4 segments, each integer). So intersection point divides each diagonal into two integer lengths. So both diagonals total i... | A convenient way to satisfy all the conditions is to take a rhombus whose four sides are equal to \(13\).
---
### 1. Conditions on the diagonals
For a rhombus the diagonals are perpendicular bisectors of each other.
If the diagonals have lengths \(d_1\) and \(d_2\), then each half‑diagonal is an integer, i.e. ... |
nemotron-math-v2-low-00000046-3a7fd3f8-c4d4-f904-b99c-32980c2ede8d | nvidia/Nemotron-Math-v2/aops | Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) that satisfy the functional equation \( f(x + y^2 - f(y)) = f(x) \) for all real numbers \( x \) and \( y \). | [
"We need f periodic shift depending on y: argument shift = y^2 - f(y). So f constant on translates by any value t = y^2 - f(y). Thus f is periodic with any period in set S = {y^2 - f(y): y∈R}. If S contains nonzero values, f constant. Could be constant function. Let f(x)=c. Then equation: c = c for all, need condit... | \[
\boxed{\;f(x)=c\ \text{for any constant }c\in\mathbb R\qquad\text{or}\qquad f(x)=x^{2}\;}
\] |
nemotron-math-v2-low-00000088-8bee3274-f88e-2470-cff5-a4c8ce1015dc | nvidia/Nemotron-Math-v2/aops | Find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that for all \( x, y \in \mathbb{R} \),
\[
(f(x+y))^2 = (f(x))^2 + 2f(yx) + (f(y))^2.
\] | [
"We need solutions. Try assume f linear: f(t)=at+b. Compute. LHS: (a(x+y)+b)^2. RHS: (ax+b)^2 +2 f(yx)+ (ay+b)^2. Compute f(yx)=a yx + b. So RHS = (ax+b)^2 +2( a yx + b) + (ay+b)^2. Expand: (a^2 x^2 +2abx + b^2) +2a yx +2b + (a^2 y^2 +2aby + b^2). Sum = a^2 (x^2 + y^2) +2ab(x+y) +2b^2 +2a yx +2b.",
"LHS: (a x + a... | \[
\boxed{\,f(x)\equiv 0,\qquad f(x)\equiv -2,\qquad f(x)=x,\qquad f(x)=x-2\,}
\]
These four functions satisfy
\((f(x+y))^{2}= (f(x))^{2}+2f(yx)+(f(y))^{2}\) for all real \(x,y\), and no other real‑valued functions do. |
nemotron-math-v2-low-00000102-9713a693-6617-4ccf-f91b-d7910126a92d | nvidia/Nemotron-Math-v2/aops | If the incentre of $\triangle ABC$ is the midpoint of the median $AD$, find $\cos A$. | [
"We have triangle ABC, D is midpoint of BC? Actually median AD from A to BC, D is midpoint of BC. Incenter I is midpoint of AD. So I is midpoint of A and D. Coordinates: let B=(0,0), C=(c,0). Then D=(c/2,0). Let A=(x,y). Incenter is intersection of angle bisectors. Coordinates formula: I = (ax_A + bx_B + cx_C)/(a+b... | \[
\boxed{-1}
\] |
nemotron-math-v2-low-00000148-781dd5fc-7b5c-ce30-fc2a-a60e6b35db1f | nvidia/Nemotron-Math-v2/aops | Find all solutions \( f: \mathbb{R} \to \mathbb{R} \) of the functional equation
\[ f(f(x) + f(y) + kxy) = xf(y) + yf(x), \quad \text{for all } x, y \in \mathbb{R}. \]
Consider the cases when \( f \) is (a) a polynomial, (b) a continuous function, and (c) an arbitrary function. | [
"We need solve. Guess linear: f(t)=at+b. Plug. Compute LHS: f(a x + b + a y + b + kxy) = a( a x + a y + kxy +2b) + b = a^2(x+y) + a k xy +2ab + b. RHS: x (a y + b) + y (a x + b)= a xy (x? wait) compute: x( a y + b) + y( a x + b) = a xy + b x + a xy + b y? Actually x*a y = a xy, y*a x = a xy, so total 2a xy + b(x+y)... | \[
\boxed{\begin{array}{l}
\text{For every real constant }k\text{ the only polynomial solutions are}\\[2mm]
\qquad f(x)\equiv 0\qquad(\text{for all }k),\\[2mm]
\qquad\text{and, when }k=2,\;f(x)= -\dfrac{x}{2}+\dfrac14 .\\[4mm]
\text{If we require }f\text{ to be continuous, the same two functions are the only
solutions ... |
nemotron-math-v2-low-00000177-d2d78a39-ac97-6fc4-66e5-0ab2b293800d | nvidia/Nemotron-Math-v2/aops | The numbers $1, 2, \dots, 1999$ are written on the board. Two players take turns choosing $a, b$ from the board, erasing them, and writing one of $ab$, $a+b$, or $a-b$. The first player wants the last number on the board to be divisible by $1999$, while the second player wants to prevent this. Determine the winner. | [
"We need parity? Invariant modulo 1999? The operation replaces a,b with one of ab, a+b, a-b. Consider product modulo 1999? The set initially contains all residues 1..1998,0? 1999 not present. Sum of all numbers mod1999 = (1999*2000/2) mod1999 = (1999*1000) mod1999 =0. So total sum ≡0. Operation: remove a,b and add ... | **Answer:** \(\boxed{\text{The first player can force a win.}}\) |
nemotron-math-v2-low-00000194-125bfa4a-b471-4592-62e4-b0c82dc7346a | nvidia/Nemotron-Math-v2/aops | Circles $C_{1}$ and $C_{2}$ are externally tangent at $M$, with the radius of $C_{2}$ being greater than the radius of $C_{1}$. Point $A$ is on $C_{2}$ and does not lie on the line joining the centers of the circles. Points $B$ and $C$ are on $C_{1}$ such that $AB$ and $AC$ are tangent to $C_{1}$. Lines $BM$ and $CM$ i... | [
"We need locus: likely a line (maybe the external homothety center). Probably D lies on fixed line through the external homothety point? The homothety center is M. The tangent at A intersect EF gives D on a fixed line through M? Might be the polar of M w.r.t. some circle? I think D lies on the line through M perpen... | **Answer.**
\[
\boxed{\text{The point }D\text{ always lies on a single fixed line – namely the polar of the
tangency point }M\text{ with respect to the larger circle }C_{2}.}
\]
**Explanation.**
Let \(O_{1}\) and \(O_{2}\) be the centres of \(C_{1}\) and \(C_{2}\), with radii \(r_{1}<r_{2}\) and
\(M\) the externa... |
nemotron-math-v2-low-00000212-4424086c-2507-f177-cba0-294d5e75f2b0 | nvidia/Nemotron-Math-v2/aops | Given two lines $\ell$ and $m$ intersecting at point $A$, and a point $M$, construct two points $B \in \ell$ and $C \in m$ such that $MC = MB$ and $M, A, B$ are collinear. | [
"We need to describe construction. Use circle centered M radius unknown? We need B on line through A and M intersect ℓ such that distance to M equals distance to C on m. So reflect M across line through A? Actually want B on line AM, C on m with MB=MC. So C is intersection of m with circle centered M radius MB. So ... | **Construction**
1. **Draw the line through \(M\) and \(A\).**
This is the unique line that must contain the points \(M,\;A,\;B\).
2. **Find the point \(B\) on \(\ell\).**
- Intersect the line \(MA\) with the given line \(\ell\).
- Let the intersection be \(B\).
(If the line \(MA\) is parallel to ... |
nemotron-math-v2-low-00000216-38f013a3-e77e-8e0c-570b-1d00b5d4c73f | nvidia/Nemotron-Math-v2/aops | Find a surjective function \( f: \mathbb{Z}^+ \to \mathbb{Z}^+ \) such that \( f(n) \geq n + (-1)^n \). | [
"We need a surjection from positive integers to positive integers satisfying f(n) >= n + (-1)^n. So for even n, (-1)^n = 1, so bound is n+1. For odd n, bound is n-1. Need to define f that hits every positive integer. Could map many n to same value. Provide example: for odd n, set f(n)=n-1 (which is even). For even ... | \[
f:\mathbb Z^{+}\longrightarrow \mathbb Z^{+},\qquad
f(n)=
\begin{cases}
1, & n=1,\\[4pt]
n-1, & n\ge 3\text{ and }n\text{ odd},\\[4pt]
n+1, & n\text{ even}.
\end{cases}
\]
---
**Verification**
*Bound.*
- If \(n\) is odd (\(n\ge3\)), then \((-1)^n=-1\) and
\[
f(n)=n-1=n+(-1)^n .
\]
- If \(n\) is even,... |
nemotron-math-v2-low-00000218-030c0246-8af9-e846-b9b2-cf4a7dc4c8e0 | nvidia/Nemotron-Math-v2/aops | Find all non-trivial solutions to the continuous function \( f: \mathbb{R} \to \mathbb{R} \) such that \( f(f(x) + 1) = f(x + 1) \). | [
"We need continuous f, non-trivial meaning not constant? Equation: f(f(x)+1)=f(x+1). Let g(x)=f(x)-c? Try identity. Suppose f(x)=x. Then LHS f(f(x)+1)=f(x+1)=x+1, RHS f(x+1)=x+1 works. So f(x)=x works.",
"Also f(x)=k constant: then LHS f(k+1)=k, RHS f(x+1)=k, works for any constant k. But non-trivial exclude cons... | \[
\boxed{\,f(x)=x\; \text{for all }x\in\mathbb{R}\,}
\]
(Aside from the trivial constant solutions \(f(x)\equiv c\) which also satisfy the equation, the only continuous non‑constant solution is the identity function.) |
nemotron-math-v2-low-00000265-2a29c97a-3ac4-7629-1973-a3a70a8b7818 | nvidia/Nemotron-Math-v2/aops | Find all functions \( f : \mathbb{R}^+ \to \mathbb{R}^+ \) such that \( f(x + f(x) + 2y) = f(2x) + y + f(y) \) for all positive real numbers \( x, y \). | [
"We look for f. Suppose linear: f(t)=at+b. Plug in. LHS: f(x+ (ax+b) +2y)= a(x+ax+b+2y)+b = a( (1+a)x +2y + b)+b = a(1+a)x +2a y + a b + b. RHS: f(2x)+ y + f(y)= a(2x)+b + y + (a y + b)= 2a x + b + y + a y + b = 2a x + a y + y +2b. Equate coefficients: x: a(1+a)=2a => if a≠0, divide a: 1+a =2 => a=1. Good. y: 2a = ... | \[
\boxed{\;f(x)=x+c\qquad (c\ge 0)\;}
\]
Indeed, substituting \(f(x)=x+c\) into
\(f(x+f(x)+2y)=f(2x)+y+f(y)\) gives
\[
\begin{aligned}
LHS &= f\bigl(x+(x+c)+2y\bigr)=f\bigl(2x+2y+c\bigr)\\
&= (2x+2y+c)+c = 2x+2y+2c,\\[4pt]
RHS &= f(2x)+y+f(y)= (2x+c)+y+(y+c)=2x+2y+2c,
\end{aligned}
\]
so the functional equa... |
nemotron-math-v2-low-00000308-2cb0ee1f-d921-eb45-f8ac-21185a0a0648 | nvidia/Nemotron-Math-v2/aops | Find all functions \( f : \mathbb{R} \rightarrow \mathbb{R} \) such that \( f(x+y) = f(x) + f(x)f\left(\frac{1}{x}\right)f(y) \) for all real numbers \( y \) and non-zero real numbers \( x \). | [
"We need functions satisfying: for all y∈R, x≠0: f(x+y)=f(x)+f(x)f(1/x)f(y). Let’s denote A(x)=f(x)f(1/x). Then equation: f(x+y)=f(x)+A(x) f(y). For each fixed nonzero x, this is Cauchy-like linear in y. Consider y variable: implies f is affine: f(t)=A(x) f(t - x)+f(x). But A depends on x. Try to find constant solu... | \[
\boxed{\,f(x)=0\quad\text{for all }x,\qquad f(x)=x\quad\text{for all }x,\qquad f(x)=-x\quad\text{for all }x\,}
\] |
nemotron-math-v2-low-00000314-b62a1095-6541-fb31-9704-88e9b6fda60a | nvidia/Nemotron-Math-v2/aops | Find all positive integers \( x_1, x_2, \ldots, x_n \) such that:
\[ x_1^3 + x_2^3 + \cdots + x_n^3 = (x_1 x_2 \cdots x_n)^2 \] | [
"We need solutions positive integers. Equation sum of cubes equals square of product. For n≥2. Try small n. For n=1: x1^3 = x1^2 => x1^3 - x1^2 = x1^2(x1-1)=0 => x1=0 or 1, positive gives x1=1. So solution (1).",
"For n=2: x^3+y^3 = (xy)^2. Try small numbers. If one is 1: 1 + y^3 = y^2 => y^3 - y^2 +1=0 => y^2(y-... | \[
\boxed{
\begin{array}{ll}
n=1:& (x_{1})=(1)\\[4pt]
n=2:& (x_{1},x_{2})=(2,2)\\[4pt]
n=3:& (x_{1},x_{2},x_{3})\text{ is any permutation of }(1,2,3)
\end{array}
}
\]
No other \(n\ge 4\) or other tuples of positive integers satisfy
\[
x_{1}^{3}+x_{2}^{3}+\dots +x_{n}^{3}=(x_{1}x_{2}\dots x_{n})^{2}.
\] |
nemotron-math-v2-low-00000328-237cc1db-af15-ad8e-4469-8994f1ca77f5 | nvidia/Nemotron-Math-v2/aops | Let \( a > 1 \). Find all functions \( f:[a,+\infty) \to [0,+\infty) \) such that
\[ f(x)f(yf(x)) = f(x+y) \]
for all \( x, y \in [a,+\infty) \). | [
"We need solutions nonnegative. Try constant? If f=c≥0. Then c * c = c => c^2=c => c=0 or1. But domain codomain nonnegative; check if works for all x,y≥a: need c=0 gives 0=0? LHS 0*0 =0 RHS 0 works. So f≡0 works. But codomain [0,+∞), ok. However need at x≥a, y≥a, argument yf(x) = y*0 =0 not in domain (needs >=a). S... | \[
\boxed{\,f(x)=1\qquad\text{for all }x\in[a,\infty)\,}
\] |
nemotron-math-v2-low-00000386-58111cfa-6c94-a076-563a-0110cb0314d3 | nvidia/Nemotron-Math-v2/aops | In a convex quadrilateral $ABCD$ with $AB = 2016$, $BC = 2017$, $CD = 2018$, and $DA = 2019$, the lines $AB$ and $DC$ intersect at $E$, and $CB$ and $DA$ intersect at $F$. Given that $CE + 5 = AF$, find the difference between $BE$ and $BF$. | [
"We need using law of sines in intersecting lines? Use complete quadrilateral with intersecting extensions. Use formula: In convex quadrilateral, intersection points E=AB∩CD, F=BC∩AD. Use lengths of sides, and relation CE+5=AF. Need BE - BF? Probably using Menelaus or similar. Might be using Stewart? Could use rati... | \[
\boxed{5}
\] |
nemotron-math-v2-low-00000400-56e2f25a-b96d-3a98-80e7-d58361317a1d | nvidia/Nemotron-Math-v2/aops | Determine the number of sequences of points \((x_1, y_1), (x_2, y_2), \dots, (x_{4570}, y_{4570})\) on the plane satisfying the following two properties:
(i) \(\{x_1, x_2, \dots, x_{4570}\} = \{1, 2, \dots, 2014\}\) and \(\{y_1, y_2, \dots, y_{4570}\} = \{1, 2, \dots, 2557\}\)
(ii) For each \(i = 1, 2, \dots, 4569\),... | [
"We need count of sequences length 4570 using all x values 1..2014 each possibly multiple times? Wait total points count =4570. Number of distinct x values =2014, y values=2557. Since lengths larger than each set, repetitions occur. Condition (i) says the multiset of x_i equals the set {1..2014}. Means each x appea... | \[
\boxed{0}
\] |
nemotron-math-v2-low-00000407-e8838526-d135-73cf-438b-7170b67684cf | nvidia/Nemotron-Math-v2/aops | Let \( f \) be a function on non-negative integers defined as follows: \( f(2n) = f(f(n)) \) and \( f(2n+1) = f(2n) + 1 \). For what non-negative integers \( k \) can \( f(0) \) equal \( 2^k \)? | [
"We need possible values of f(0)=2^k given constraints. Try define? Let a = f(0). Use recurrence. Compute some values: Let n=0 => f(0)=f(f(0)) => a = f(a). Also f(1)=f(0)+1 = a+1.",
"Now consider n=a? not needed. Need consistency. Likely only powers of 2? Let's test small. Suppose a=1 (2^0). Then f(1)=2. Also f(1... | The functional equations force a contradiction for every possible value of
\[
f(0)=2^{k}\qquad(k\in \mathbb N_{0}),
\]
so there is **no** non‑negative integer \(k\) for which such an \(f\) can exist.
\[
\boxed{\text{No such }k}
\] |
nemotron-math-v2-low-00000417-d551167e-45c0-abc4-e792-55a5fb7d195e | nvidia/Nemotron-Math-v2/aops | Solve the equation \(3^a + 7^b = n^2\) for natural numbers \(a\), \(b\), and \(n\). | [
"We need solutions. Likely small. Try parity: 3^a mod4 = (3 mod4)^a => if a odd =>3 mod4, even=>1. 7^b mod4: 7≡3 => similar. Sum mod4 possibilities: 1+1=2, 1+3=0,3+1=0,3+3=2. Square mod4 is 0 or1. So sum cannot be 2 mod4, so need sum ≡0 or1. So cannot have both terms ≡1 (i.e., a,b even) -> sum 2 not allowed. So at ... | \[
\boxed{(a,b,n)=(2,\;1,\;4)}
\] |
nemotron-math-v2-low-00000430-1c0700e6-fc16-7da0-ee97-fde0e89fe0c6 | nvidia/Nemotron-Math-v2/aops | Let \(\triangle ABC\) be given with circumradius \(R\). Let \(K_1\) and \(K_2\) be the circles which pass through \(C\) and are tangent to \(AB\) at \(A\) and \(B\) respectively. Let \(K\) be the circle of radius \(r\) which is tangent to \(K_1\), \(K_2\), and \(AB\).
(a) Express \(r\) in terms of \(a\), \(b\), \(c\), ... | [
"We need geometry. Likely r = (abc)/(a+b+c+? ). Actually circles tangent to AB at A and B through C => K1 has center at some point on line through A perpendicular to AB? Since tangent at A, radius line is perpendicular to AB through A. So K1 center is somewhere on line through A perpendicular to AB, distance from A... | **Answer**
\[
\boxed{\displaystyle
\begin{aligned}
\text{(a)}\qquad
r&=\frac{R\,\sin\frac A2\,\sin\frac B2}{\sin\frac C2}\;,\\[4pt]
\text{(b)}\qquad
\angle C&=90^{\circ}\;.
\end{aligned}}
\] |
nemotron-math-v2-low-00000439-1462d9a5-446c-5ea7-18e2-d4e695c80f6f | nvidia/Nemotron-Math-v2/aops | Given a set of points in the plane, define a "fit" rectangle as a (not necessarily nondegenerate) rectangle that has a diagonal whose endpoints are two of these points. Find the least number that represents the maximum number of points from this set that can be bounded by a fit rectangle (or placed on its sides). | [
"We need answer? Problem unclear: likely answer is 4? Max points bounded by rectangle defined by two points as opposite corners: all points with x between and y between. So maximum possible is number of points that lie in axis-aligned rectangle? But rectangle can be any orientation since diagonal endpoints arbitrar... | The rectangle is defined by two points of the set that form a diagonal.
No matter how the points are arranged, we can always choose those two points as the
end‑points of a diagonal; the rectangle they determine will certainly contain those
two points (they lie at its opposite corners).
Thus, for any configuration ... |
nemotron-math-v2-low-00000453-8e610a37-6448-2ce0-ea48-72f6eb17ea99 | nvidia/Nemotron-Math-v2/aops | Given the equation $z(x+y)=z+xy$, find the value of $x+y+z$. | [
"We solve equation: z(x+y)=z+xy. Want x+y+z in terms maybe constant? Rearr: zx+zy = z + xy. Bring all: zx+zy - xy - z =0 => (zx - xy) + (zy - z)=0 => x(z - y) + z(y -1)=0. Not enough info; perhaps integer solutions? Maybe assume x,y,z are numbers satisfying? Could be symmetric? Solve for sum? Try express z in terms... | First, suppose the three numbers are equal:
\[
x=y=z.
\]
Substituting \(x=y=z\) into the given equation
\[
z(x+y)=z+xy
\]
gives
\[
z( x+x)=z+x^2\quad\Longrightarrow\quad 2z^2 = z+z^2 .
\]
Thus
\[
z^2 - z = 0 \quad\Longrightarrow\quad z(z-1)=0,
\]
so \(z=0\) or \(z=1\).
If \(z=0\) then \(x=y=0\) and \(... |
nemotron-math-v2-low-00000515-4e2284de-1c84-bd32-4d12-bf63f7c5e64b | nvidia/Nemotron-Math-v2/aops | Find a natural number \( n \) that is divisible by 17 and satisfies the following conditions:
- \( \frac{n}{2} \) is a power of 2.
- \( \frac{n}{3} \) is a power of 3.
- \( \frac{n}{4} \) is a power of 5. | [
"We need n divisible by 17, also n/2 = 2^a, n/3 = 3^b, n/4 =5^c with integers a,b,c >=0. So n = 2^{a+1} = 3^{b+1} = 5^{c+2}. Actually n = 2 * 2^a = 2^{a+1}. Similarly n = 3 * 3^b = 3^{b+1}. And n = 4 * 5^c = 2^2 *5^c = 2^{2} *5^c.",
"Thus n must be simultaneously a power of 2, a power of 3, and 2^2*5^c. The only ... | Let
\[
\frac{n}{2}=2^{a},\qquad \frac{n}{3}=3^{b},\qquad \frac{n}{4}=5^{c}
\]
for some non‑negative integers \(a,b,c\).
Then
\[
n = 2\cdot 2^{a}=2^{a+1},\qquad
n = 3\cdot 3^{b}=3^{b+1},\qquad
n = 4\cdot 5^{c}=2^{2}\,5^{c}.
\]
Hence \(n\) must simultaneously be a power of \(2\), a power of \(3\), and also of... |
nemotron-math-v2-low-00000551-3b3a1a87-4cd0-c943-2a86-778cedfd614c | nvidia/Nemotron-Math-v2/aops | Find all polynomials $P(x) \in \mathbb{R}[x]$ such that $P(x)^n = P(x^n)$ for $n \in \mathbb{Z}$. | [
"We need polynomials satisfying P(x)^n = P(x^n) for all integers n? Probably all n≥1. Consider degree d. Then left deg = n d, right deg = d * n? Actually P(x^n) has degree d*n. So degrees match fine. Compare leading coefficient a. LHS leading coeff a^n x^{nd}. RHS leading coeff a (x^n)^d = a x^{nd}. So a^n = a => a... | \[
\boxed{\;P(x)=0,\qquad P(x)=1,\qquad P(x)=x^{k}\;(k\in\mathbb{N}_{0})\;}
\]
These are the only real‑coefficient polynomials that satisfy
\[
P(x)^{\,n}=P\bigl(x^{n}\bigr)\qquad\text{for every integer }n\ge 1 .
\] |
nemotron-math-v2-low-00000566-3d67c76e-3f74-a1b2-a7c8-69ce4dfc7e3f | nvidia/Nemotron-Math-v2/aops | Evaluate \( \sin\alpha \cdot \sin2\alpha \cdot \sin4\alpha \). | [
"We need expression? Probably simplify product formula: sinα sin2α sin4α = (1/4) sinα sin2α *2? Use identity: sin x sin 2x = (1/2)(cos(x-2x)-cos(x+2x)) = (1/2)(cos(-x)-cos3x)= (1/2)(cos x - cos3x). Then multiply by sin4α: (1/2)(cosα - cos3α) sin4α. Then use product to sum: cosA sinB = (1/2)[sin(B+A)+ sin(B-A)]. So ... | \[
\sin\alpha\;\sin2\alpha\;\sin4\alpha
\]
\[
\begin{aligned}
\sin\alpha\sin2\alpha
&=\tfrac12\bigl(\cos(\alpha-2\alpha)-\cos(\alpha+2\alpha)\bigr) \\
&=\tfrac12\bigl(\cos\alpha-\cos3\alpha\bigr)
\end{aligned}
\]
Hence
\[
\sin\alpha\sin2\alpha\sin4\alpha
=\tfrac12\bigl(\cos\alpha-\cos3\alpha\bigr)\sin4\a... |
nemotron-math-v2-low-00000586-76a34c5e-708a-a563-2542-3e0c46063400 | nvidia/Nemotron-Math-v2/aops | Find all values of the parameter \( a \) such that the equations \( x^2 - ax + 1 = 0 \) and \( x^2 - x + a = 0 \) have at least one common solution. For these values of \( a \), find the common solution of these equations (real or imaginary). | [
"We need common root r satisfies both equations: r^2 - a r +1 =0 and r^2 - r + a=0. Subtract: (r^2 cancels) -a r +1 - (- r + a)= -a r +1 + r - a =0 => r(1 - a) + (1 - a)=0 => (1 - a)(r +1)=0.",
"Thus either a=1 or r = -1.",
"Case 1: a=1. Then both equations become: x^2 - x +1=0 and x^2 - x +1=0 same, roots are ... | Let a common root be \(r\). Then
\[
\begin{cases}
r^{2}-ar+1=0,\\[2mm]
r^{2}-r+a=0 .
\end{cases}
\]
Subtract the second equation from the first:
\[
\bigl(r^{2}-ar+1\bigr)-\bigl(r^{2}-r+a\bigr)=0
\Longrightarrow -ar+1+r-a=0
\Longrightarrow (1-a)(r+1)=0 .
\]
Hence either
\[
\boxed{a=1}\qquad\text{or}\qquad\boxed... |
nemotron-math-v2-low-00000591-d56e78ae-be7f-1c09-d5d0-d145acd047c8 | nvidia/Nemotron-Math-v2/aops | For a triangle with a fixed area \( A \), what are the maximum and minimum possible perimeters? | [
"We need max and min perimeter given fixed area. No side length constraints, shape can be degenerate? Minimum perimeter for given area is equilateral triangle. So P_min = 3 * side, area = (sqrt3/4) s^2 = A → s = 2√(A/√3). Then perimeter = 3s = 6√(A/√3)=6√(A)/3^{1/4}. Simplify: P_min = 6 A^{1/2} / 3^{1/4}. Could wri... | Let a triangle have side lengths \(a,b,c\) and area \(A\) (fixed).
Denote its perimeter by \(P=a+b+c\).
---
### 1. Lower bound for the perimeter
For a triangle with a given area the *isoperimetric inequality* for polygons tells us that among all triangles the **equilateral triangle** has the smallest perimete... |
nemotron-math-v2-low-00000618-b5cb7cce-c086-3df3-b454-d8dc46d2c706 | nvidia/Nemotron-Math-v2/aops | Find a function \( f: \mathbb{R} \to \mathbb{R}^+ \) such that:
\[ f(x)^2 = 1 + x f(x+1) \] | [
"We need any function satisfying functional equation. Try constant? Let f = c>0. Then c^2 =1 + x c => c^2 - x c -1 =0 for all x impossible.",
"Maybe define recursively? Solve for f(x): f(x)= (1 + x f(x+1))^{1/2}. Hard.",
"Maybe look for linear: f(x)=a x + b. Compute: (a x + b)^2 =1 + x (a (x+1)+b) =1 + x (a x +... | \[
\boxed{f(x)=x+1}
\]
Indeed, substituting \(f(x)=x+1\) into the functional equation gives
\[
\begin{aligned}
f(x)^{2} &= (x+1)^{2}=x^{2}+2x+1,\\[2mm]
1+x\,f(x+1)&=1+x\,(x+2)=1+x^{2}+2x=x^{2}+2x+1,
\end{aligned}
\]
so the equality \(f(x)^{2}=1+x\,f(x+1)\) holds for every real \(x\).
Since \(f(x)=x+1>0\) for all... |
nemotron-math-v2-low-00000622-d73f7f9d-a963-b897-2dfa-b34fdb7ed51a | nvidia/Nemotron-Math-v2/aops | Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) such that for all \( x, y \in \mathbb{R} \),
\[ f(3x - f(y)) = f(f(y)) - 6xf(y) + f(3x). \] | [
"We need find f. Suppose polynomial. Let’s try assume f is quadratic: f(t)=at^2+bt+c. Compute? Might be linear. Try f(t)=kt + m. Then LHS: f(3x - (ky+m)) = k(3x - ky - m)+m = 3kx - k^2 y - km + m. RHS: f(f(y)) = k(ky+m)+m = k^2 y + km + m. Then -6x f(y)= -6x(ky+m)= -6kxy -6mx. + f(3x)= k(3x)+m=3kx+m. Sum RHS: (k^2 ... | \[
\boxed{\,f(x)=0\quad\text{for all }x\in\mathbb R\quad\text{or}\quad f(x)=x^{2}\quad\text{for all }x\in\mathbb R\,}
\] |
nemotron-math-v2-low-00000673-a3415c93-69c1-aebe-034b-ce4bfda0038e | nvidia/Nemotron-Math-v2/aops | Find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that for all \( x, y \in \mathbb{R} \),
\[
(f(x+y))^2 = (f(x))^2 + 2f(yx) + (f(y))^2.
\] | [
"We need solutions. Try assume f is linear: f(t)=at+b. Compute LHS: (a(x+y)+b)^2 = a^2(x+y)^2+2ab(x+y)+b^2. RHS: (ax+b)^2 +2 f(yx)+ (ay+b)^2 = a^2x^2+2abx+b^2 +2( a(yx)+b ) + a^2 y^2+2aby+b^2. Sum = a^2(x^2+y^2)+2ab(x+y)+2b^2 +2a xy+2b. Compare with LHS: a^2(x^2+2xy+y^2)+2ab(x+y)+b^2. So LHS = a^2x^2+2a^2 xy + a^2 ... | \[
\boxed{\;f(x)=0,\qquad f(x)=-2,\qquad f(x)=x,\qquad f(x)=x-2\; }
\] |
nemotron-math-v2-low-00000692-e1b16cae-4c87-83db-5092-cf0a6d3ec127 | nvidia/Nemotron-Math-v2/aops | For \(a \neq 0\) and \(a, b, c \in \mathbb{R}\), find the function \(f: \mathbb{R} \rightarrow \mathbb{R}\) such that:
\[ f(x - f(y)) = f(x) + f(f(y)) - axf(y) - bf(y) - c \]
for all \(x, y \in \mathbb{R}\). | [
"We need find f linear? Assume f(t)=pt+q. Compute. LHS: f(x - (p y+ q)) = p(x - p y - q)+q = p x - p^2 y - p q + q. RHS: f(x)= p x+q. f(f(y)) = f(p y+ q)= p(p y+ q)+q = p^2 y + p q + q. Then RHS sum: (p x+q)+(p^2 y + p q + q) - a x (p y+ q) - b (p y+ q) - c.",
"Combine: = p x + q + p^2 y + p q + q - a p x y - a q... | \[
\boxed{\,f(x)\equiv 0\text{ for all }x\in\mathbb R\quad\text{and this is possible only when }c=0\,}
\] |
nemotron-math-v2-low-00000708-43ae5aca-0849-a3ea-9b6d-de273f7c65ec | nvidia/Nemotron-Math-v2/aops | How can the expression $x(a-b)^2 + y(b-c)^2 + z(c-a)^2$ be converted to the form $m(a-b)^2 + n(a-c)(b-c)$? | [
"We need to express given sum in that form. Compute:",
"x(a-b)^2 + y(b-c)^2 + z(c-a)^2.",
"We want = m(a-b)^2 + n(a-c)(b-c).",
"Expand both sides.",
"Let’s expand LHS:",
"(a-b)^2 = a^2 -2ab + b^2\n(b-c)^2 = b^2 -2bc + c^2\n(c-a)^2 = c^2 -2ca + a^2",
"LHS = x(a^2 -2ab + b^2) + y(b^2 -2bc + c^2) + z(c^2 -... | \[
\begin{aligned}
x(a-b)^{2}+y(b-c)^{2}+z(c-a)^{2}
&=m(a-b)^{2}+n(a-c)(b-c)\\[4pt]
\Longleftrightarrow\;&
\begin{cases}
m = x+z,\\
m = x+y,\\
n = y+z,\\
-2m+n = -2x,\\
-n = -2z,\\
-n = -2y.
\end{cases}
\end{aligned}
\]
From the last two equations we obtain \(n=2z\) and \(n=2y\); hence
\[
\boxed{y=z}.
\]
With \(y=... |
nemotron-math-v2-low-00000715-1ba10a11-e587-239f-e0a1-b98c1272ad6e | nvidia/Nemotron-Math-v2/aops | Find all continuous functions \( f: \mathbb{R} \to \mathbb{R} \) satisfying the equation:
\[ f(x+y) + f(xy) = f(x) + f(y) + f(xy+1) \] | [
"We need solve functional equation continuous. Try to find form. Let’s attempt polynomial. Suppose f is linear: f(t)=at+b. Plug in: a(x+y)+b + a(xy)+b = a x + b + a y + b + a(xy+1)+b. LHS = a(x+y+xy)+2b. RHS = a(x+y+xy+1)+3b. Compare: coefficients same, constants: 2b = 3b + a*1? Wait RHS constant part: a*1 +3b. So ... | \[
\boxed{\,f(x)=ax^{2}+bx+c\qquad\text{with}\qquad a+b+c=0\;(a,b,c\in\mathbb R)\,}
\]
Indeed, every quadratic polynomial whose coefficients satisfy \(a+b+c=0\) fulfills
\[
f(x+y)+f(xy)=f(x)+f(y)+f(xy+1)
\]
and, by continuity, these are the only real‑valued functions on \(\mathbb R\) that do. (The linear case \(a=... |
nemotron-math-v2-low-00000717-37c009b6-1668-71c4-c3f6-d9174f9d13e8 | nvidia/Nemotron-Math-v2/aops | Given the number \( \overline{a_1a_2a_3a_4a_5a_6a_7\ldots} = 2009^{2008^{2007^{2006^{2005^{2004^{2003^{2002^{2001^{2000}}}}}}}}} \), find the digit \( a_{2009^{2009}} \) in the decimal representation of this number. | [
"We need the digit at position 2009^{2009} from left? Probably decimal expansion of integer huge. a1 is most significant digit. They ask digit at that far position. Need number of digits? The exponent huge, but number is 2009^{ huge}. Number of digits = floor(log10(2009^E)) +1 = floor(E*log10 2009)+1. a_n correspon... | The digit \(a_{2009^{2009}}\) is
\[
\boxed{0}
\] |
nemotron-math-v2-low-00000740-8e54dafe-cf6e-67c3-cab9-2a20db0f3dfa | nvidia/Nemotron-Math-v2/aops | Find all triples $(p, n, k)$ such that $p^n + n = (n + 1)^k$, where $n, k \in \mathbb{Z}^+$ and $p$ is a Fermat prime number. | [
"We need solutions with p a Fermat prime: known Fermat primes: 3,5,17,257,65537. Try small n.",
"Equation: p^n + n = (n+1)^k.",
"Check n=1: p^1+1 = 2^k => p+1 = 2^k. So p = 2^k -1 must be Fermat prime. Solve for k. For k=1: p=1 (no). k=2: p=3 (Fermat prime). k=3: p=7 not Fermat. k=4: p=15 not. k=5: p=31 not. k=... | \[
\boxed{(p,n,k)=(3,1,2)\quad\text{or}\quad(p,n,k)=(5,2,3)}
\] |
nemotron-math-v2-low-00000760-1781ae77-8919-446a-fc5e-b697055b025d | nvidia/Nemotron-Math-v2/aops | Let \(a\), \(b\), and \(x\) be positive real numbers distinct from one. Determine the condition under which the equation \[4(\log_a x)^2 + 3(\log_b x)^2 = 8(\log_a x)(\log_b x)\] holds. | [
"We need condition relating a,b,x. Let u=log_a x, v=log_b x. Equation: 4u^2 +3v^2 =8uv. Rearr: 4u^2 -8uv +3v^2 =0. Consider as quadratic in u: 4u^2 -8v u +3v^2 =0. Solve: u = [8v ± sqrt(64v^2 -48v^2)]/(8) = [8v ± sqrt(16v^2)]/8 = [8v ±4v]/8 = v*(8±4)/8.",
"Thus u = v*(12/8)= (3/2)v or u = v*(4/8)= (1/2)v.",
"So... | Let
\[
u=\log_a x,\qquad v=\log_b x\qquad (x>0,\;x\neq 1,\;a,b>0,\;a,b\neq 1)
\]
The given equation becomes
\[
4u^{2}+3v^{2}=8uv .
\]
Treat it as a quadratic in \(u\):
\[
4u^{2}-8vu+3v^{2}=0 .
\]
Solve for \(u\):
\[
u=\frac{8v\pm\sqrt{(8v)^2-4\cdot4\cdot3v^{2}}}{2\cdot4}
=\frac{8v\pm\sqrt{64v^{2}-48v^{2}}... |
nemotron-math-v2-low-00000769-4e311de9-a40b-f9c0-4e32-3802c8e5d50f | nvidia/Nemotron-Math-v2/aops | Using the letters k, a, l, a, m, a, z, o, o, how many words can be written such that the letter 'o' appears before the letter 'k'? | [
"We have letters: k, a, l, a, m, a, z, o, o. That's 9 letters with repeats: a appears 3 times, o appears 2 times, others distinct (k,l,m,z). Total permutations = 9! / (3! 2!) . We need those where any o appears before k. Since there are two o's, condition: at least one o comes before k? Actually phrase \"the letter... | The multiset of letters is
\[
\{k, a, a, a, l, m, z, o, o\}
\]
There are \(9\) letters with three a’s and two o’s.
The total number of distinct permutations (words) is
\[
\frac{9!}{3!\,2!}= \frac{362880}{6\cdot 2}=30240 .
\]
---
### Condition on the letters **o** and **k**
We need the words in which **at le... |
nemotron-math-v2-low-00000773-76a5598d-7af1-6a74-374d-142d79335c9b | nvidia/Nemotron-Math-v2/aops | Find the volume of an octagonal pyramid with curved side lengths, where the radius of the base is 20 cm and the height from the center of the base to the apex is 20 cm. Assume the arcs are like those of a circle. | [
"We need volume of a pyramid with base an octagon? \"curved side lengths\" meaning base is an octagon formed by arcs of a circle radius 20? So base area is area of regular octagon inscribed in circle radius 20? Actually radius of base =20 cm likely distance from center to vertices? If arcs like those of a circle, t... | The pyramid’s base is not a straight‑edged octagon; each of the eight “sides’’ is an arc of a circle.
Thus the base is simply the region enclosed by a full circle of radius
\[
R = 20\text{ cm}.
\]
---
### 1. Area of the base
\[
A_{\text{base}} = \pi R^{2}= \pi(20)^{2}=400\pi\ \text{cm}^{2}.
\]
---
### 2. ... |
nemotron-math-v2-low-00000774-7e47e8fd-804e-a9d7-f9ae-06621ab9aea4 | nvidia/Nemotron-Math-v2/aops | Two intelligent players (A and B, with A playing first) take turns filling numbers $1, 2, \dots, 9$ in a $3 \times 3$ grid without repetition. A wins if the determinant of the grid is negative, B wins if the determinant is positive, and it's a tie if the determinant is zero. Determine who has a winning strategy. | [
"We need to see game outcome. Determinant of 3x3 matrix with entries 1-9 permutation. Sum of all entries fixed =45. Determinant sign? Probably second player can force zero? I suspect B has winning strategy? Need reasoning: determinant is linear in each row. Might be symmetry: swapping two rows changes sign. Since b... | \[
\boxed{\text{Player B has a winning strategy.}}
\] |
nemotron-math-v2-low-00000783-d633786a-ff33-ecae-8a30-f18f21e30e89 | nvidia/Nemotron-Math-v2/aops | Find all functions \( f \) such that
\[ f(f(x) + y) = x + f(y) \]
for all real numbers \( x \) and \( y \). | [
"We look for bijective. Let P(x,y). Set y = 0: f(f(x)) = x + f(0). (1)",
"Thus f is bijective. Let c = f(0). Then from (1) f is involution up to shift.",
"We can try to find linear solutions: assume f(x)=ax+b. Plug in: f(a x + b + y)= a(ax+b+y)+b = a^2 x + a y + a b + b. RHS: x + a y + b. Equate coefficients: f... | \[
\boxed{\,f(x)=x\quad\text{or}\quad f(x)=-x\; \text{for all }x\in\mathbb R\,}
\] |
nemotron-math-v2-low-00000818-b3514462-087e-dd10-26e6-900dc47022f2 | nvidia/Nemotron-Math-v2/aops | Find rational numbers \(\frac{a}{b}\) such that \(\frac{2019}{2020} < \frac{a}{b} < \frac{2021}{2020}\). | [
"We need rational numbers between those two fractions. Many exist. Probably they want simplest? Maybe find all with denominator 2020? Then a between 2019 and 2021 so a=2020 gives 2020/2020=1 which is between. So answer a/b = 1. Also any fraction reduced. So answer: any rational between. Provide example: 1, 2021/202... | \[
\frac{2019}{2020}<\frac{a}{b}<\frac{2021}{2020}
\qquad(\;a,b\in\mathbb Z,\;b>0\;).
\]
Because the two endpoints differ by only
\[
\frac{2021}{2020}-\frac{2019}{2020}=\frac{2}{2020}=\frac1{1010},
\]
there are infinitely many rational numbers lying between them.
A convenient way to exhibit **all** such rational... |
nemotron-math-v2-low-00000825-50bd695e-0166-01cf-6ef4-c94a68ce797b | nvidia/Nemotron-Math-v2/aops | Given the equation \( A + B = A \left( \frac{B(A^2) + 10}{X} \right) + B \left( \frac{B-1}{Y} \right) \), find all possible values of \( X \) and \( Y \). | [
"We need solve for X,Y given arbitrary A,B? Probably identity for all A,B so coefficients match. Write equation: A+B = A*( (B A^2 +10)/X ) + B*( (B-1)/Y ). Expand: = A*(B A^2 +10)/X + B*(B-1)/Y.",
"Treat as polynomial in A,B. Must hold for all A,B. So compare terms.",
"Compute expression: term A*(B A^2)/X = (B ... | First, rewrite the given equation
\[
A+B = A\!\left(\frac{B A^{2}+10}{X}\right)+B\!\left(\frac{B-1}{Y}\right)
\]
and separate the two terms on the right‑hand side:
\[
A\left(\frac{B A^{2}+10}{X}\right) \quad\text{and}\quad
B\left(\frac{B-1}{Y}\right).
\]
For the left‑hand side to be exactly \(A+B\), each of thes... |
nemotron-math-v2-low-00000838-d18185f7-833a-115a-1db0-9e1d696f6e2e | nvidia/Nemotron-Math-v2/aops | Let \( S = \{1, 2, 3, \ldots, n\} \), where \( n \) is an odd number. Find the parity of the number of permutations \( \sigma : S \to S \) such that the sequence defined by \( a(i) = |\sigma(i) - i| \) is monotonous. | [
"We need parity (even/odd) of count of permutations where absolute differences monotonic (non-decreasing or non-increasing?). \"monotonous\" ambiguous. Likely monotone (non-decreasing). Count permutations where |σ(i)-i| is monotonic sequence. For odd n. Need parity. Probably odd. Let's test small n.",
"n=1: only ... | **Answer**
\[
\boxed{\text{odd}}
\]
**Explanation (sketch).**
For an odd size set \(S=\{1,\dots ,n\}\) the number of permutations \(\sigma\) for which the sequence
\[
a(i)=|\sigma(i)-i|,\qquad i=1,\dots ,n,
\]
is monotone (i.e. non‑decreasing) is always odd.
The parity can be proved by constructing a sign‑rev... |
nemotron-math-v2-low-00000845-6b3d0591-772d-715d-5d8e-7d49afbd459a | nvidia/Nemotron-Math-v2/aops | Find all positive integer pairs \(a, b\) such that \(a^b - b^a = n^2\), where \(n\) is an integer. | [
"We need solutions to a^b - b^a = perfect square. Probably only small. Try a=b gives zero square. So all a=b works, gives n=0. Positive integer pairs. Also maybe (2,1): 2^1-1^2=2-1=1 => n=±1. (3,1):3-1=2 not square. (3,2):3^2-2^3=9-8=1 square. (4,2):4^2-2^4=16-16=0 square. (4,3):4^3-3^4=64-81=-17 no. (5,2):25-32=-7... | \[
\boxed{\begin{aligned}
&\text{1. }a=b\qquad (\text{any positive integer }a,\;n=0);\\[2mm]
&\text{2. }b=1,\;a=k^{2}+1\qquad (k\in\mathbb Z_{>0},\;n=k);\\[2mm]
&\text{3. }(a,b)=(3,2)\qquad (n=1);\\[2mm]
&\text{4. }(a,b)=(4,2)\qquad (n=0).
\end{aligned}}
\]
These are all the positive integer pairs \((a,b)\) for which ... |
nemotron-math-v2-low-00000866-88443268-bfca-4020-cfab-85ec03022cd2 | nvidia/Nemotron-Math-v2/aops | Find a neat formula for $\sum^{n-1}_{k=0}{(z^{2k}-p)^{2m}}$, where $m,n\in \mathbb{N}$. | [
"We need formula: sum_{k=0}^{n-1} (z^{2k} - p)^{2m}. Use roots of unity: z^{2k}= (z^2)^k. Let ω = z^2. Sum_{k=0}^{n-1} (ω^k - p)^{2m}. Expand via binomial: = Σ_{j=0}^{2m} binom{2m}{j} (-p)^{2m-j} Σ_{k=0}^{n-1} ω^{kj}. The inner sum is n if ω^{j}=1 i.e. ω^j=1 mod? ω is arbitrary complex, not root of unity generally.... | Let
\[
\omega = z^{2}\qquad (z\in\mathbb C),\qquad
S:=\sum_{k=0}^{n-1}\bigl(z^{2k}-p\bigr)^{2m}
=\sum_{k=0}^{n-1}\bigl(\omega^{k}-p\bigr)^{2m}.
\]
---
### 1. Expand the power
\[
(\omega^{k}-p)^{2m}= \sum_{j=0}^{2m}\binom{2m}{j}(\omega^{k})^{j}(-p)^{\,2m-j}.
\]
Hence
\[
S =\sum_{j=0}^{2m}\binom{2m}{j}... |
nemotron-math-v2-low-00000887-c8e26fa5-305c-4c52-6256-059b3b5f0d1c | nvidia/Nemotron-Math-v2/aops | Find all polynomials with integer coefficients such that for all prime numbers \( p \), \( f(p) \) divides \( 2^p - 2 \). | [
"We need polynomials f with integer coeffs such that for every prime p, f(p) | 2^p -2. Note 2^p-2 = 2(2^{p-1}-1). For prime p, by Fermat 2^{p-1}≡1 mod p unless p=2. So p divides 2^{p-1}-1, so p divides 2^p-2. Also for any divisor of p maybe.",
"Thus requirement that f(p) is a divisor of that number, which is mult... | \[
\boxed{\;f(x)=\pm 1,\ \pm 2,\ \pm x\;}
\]
These are the only polynomials with integer coefficients that satisfy
\[
f(p)\mid 2^{p}-2\qquad\text{for every prime }p .
\] |
nemotron-math-v2-low-00000910-32b7fdc7-5728-2150-3b36-62a02fc19c8d | nvidia/Nemotron-Math-v2/aops | If the 9th, 13th, and 15th terms of an arithmetic progression are the first three terms of a geometric progression where \(a = 80(1 - r)\), find the first term of the arithmetic progression and the first term of the geometric progression. | [
"We have AP: let first term A, common difference d. Then term n = A + (n-1)d. So term9 = A+8d, term13 = A+12d, term15 = A+14d. These three are first three terms of GP: so ratio r = (term13)/(term9) = (term15)/(term13). So (A+12d)/(A+8d) = (A+14d)/(A+12d). Cross-multiply: (A+12d)^2 = (A+8d)(A+14d). Expand: (A^2 +24A... | The arithmetic progression has first term \(A\) and common difference \(d\).
\[
\begin{aligned}
T_9 &= A+8d,\\
T_{13} &= A+12d,\\
T_{15} &= A+14d.
\end{aligned}
\]
These three terms are the first three terms of a geometric progression, so
\[
\frac{A+12d}{A+8d}= \frac{A+14d}{A+12d}=r .
\]
Equating the two ratios:
\... |
nemotron-math-v2-low-00000921-250e52cf-03d2-2b1a-afdf-3c5046c040bf | nvidia/Nemotron-Math-v2/aops | Let \( a_{1} = 2 \) and \( a_{2} = 3 \). For each \( n \geq 3 \), \( a_{n} \) is determined by a coin toss: \( a_{n} = 2a_{n-2} \) if a head occurs, and \( a_{n} = 3a_{n-1} - 2a_{n-2} \) otherwise. Find the probability that there exists an \( n \) for which \( 1600 < a_{n} < 2000 \). | [
"We need probability over infinite sequence of coin tosses that some term falls in interval. Need to analyze possible values. Recurrence: two possibilities depending on head (H) or tail (T). Let's define state as pair (a_{n-1}, a_{n})? But recurrence uses previous two. Simulate possible values? Might be linear comb... | \[
\boxed{0}
\] |
nemotron-math-v2-low-00000942-c6a57191-7406-081a-a094-cb607d3ee814 | nvidia/Nemotron-Math-v2/aops | In a triangle \(ABC\), given that \((\sin A - \cos A)(\sin B - \cos B)(\sin C - \cos C) = 1\), find the value of \(\cot A + \cot B + \cot C\). | [
"We have product =1. Use identity sin - cos = sqrt2 sin(A-45°). So product = (√2)^3 sin(A-π/4) sin(B-π/4) sin(C-π/4)=1. So 2√2 * product sin =1 => product sin = 1/(2√2). But A+B+C=π. Let x=A-π/4, y=B-π/4, z=C-π/4. Then x+y+z = π - 3π/4 = π/4. So we have sin x sin y sin z = 1/(2√2). Need cot A+cot B+cot C = sum cot(... | \[
\boxed{1}
\] |
nemotron-math-v2-low-00000953-78b1e58a-9d7a-a994-df33-4c47d576330a | nvidia/Nemotron-Math-v2/aops | Assume $1+2+\cdots +n$ is a three-digit number with all digits equal. Then $n$ is divisible by which of the following numbers? | [
"We need sum = n(n+1)/2 is a three-digit repdigit: 111,222,...,999. Find n integer. Compute n(n+1)=2*repdigit. Try each repdigit.",
"Let S=111 => 2S=222. Solve n^2 + n -222 =0. Discriminant D=1+888=889 not perfect square. No.",
"222 => 2S=444. n^2 + n -444=0. D=1+1776=1777 not square.",
"333 => 666. D=1+2664=... | The sum of the first \(n\) positive integers is
\[
S=\frac{n(n+1)}{2}.
\]
We are told that \(S\) is a three‑digit number whose three digits are all the same, i.e.
\[
S\in\{111,222,333,444,555,666,777,888,999\}.
\]
Hence
\[
\frac{n(n+1)}{2}=k\qquad(k=111,222,\dots ,999)
\]
or
\[
n^{2}+n-2k=0 .
\]
For eac... |
nemotron-math-v2-low-00000955-5cd41620-5761-a0f2-a96c-41696f5b982a | nvidia/Nemotron-Math-v2/aops | Given \(x, y, z \geq 0\), find the value of \(\frac{1}{9}(x + y + z)^2(xy + yz + zx)\). | [
"We need expression value? Perhaps it's constant? Maybe given condition missing? Probably using inequality? Might be minimum? Probably find minimum value. For nonnegative, minimize (1/9)(sum)^2 * sum pair. Use AM-GM: Let s = x+y+z, p = xy+yz+zx. Need min of (s^2 p)/9. For nonnegatives, p ≤ s^2/3 (since (x+y+z)^2 = ... | The expression
\[
\frac{1}{9}(x+y+z)^{2}(xy+yz+zx)
\]
is non‑negative for all \(x,y,z\ge 0\).
To see its smallest possible value, notice that if any two of the variables are zero, say \(y=z=0\), then
\[
\frac{1}{9}(x+0+0)^{2}(x\cdot 0+0\cdot 0+0\cdot x)=\frac{1}{9}x^{2}\cdot 0=0 .
\]
Since the expression cann... |
nemotron-math-v2-low-00000960-2e890b5b-c1c9-f467-8375-d9d74ec45dbe | nvidia/Nemotron-Math-v2/aops | What is the greatest number of rooks that can be placed on an $n \times n$ chessboard such that each rook attacks an even number of other rooks? A rook is said to attack another rook if they are on the same row or column with no other rooks between them. | [
"We need max rooks with each rook having even degree in rook graph with adjacency if visible. Likely all squares: n^2 works? But interior rooks see many. Need each rook's visible count even. Probably max is n^2 for even n, and n^2 -1 for odd? Let's think: parity of total sum of degrees = 2*edges, even. Each rook ev... | \[
\boxed{\begin{cases}
n^{2}, &\text{if } n \text{ is even},\\[4pt]
n^{2}-1, &\text{if } n \text{ is odd}.
\end{cases}}
\] |
nemotron-math-v2-low-00001045-3b4bbfb6-6f7e-85ff-fdca-2dfd0fbfb9aa | nvidia/Nemotron-Math-v2/aops | Find all injective functions \( f \colon \mathbb{N} \to \mathbb{N} \) such that the following conditions hold for all sufficiently large \( n \), \( a \), and \( b \):
1. \(\frac{f(n)}{n} \le \ln(n)^{\frac{d(n)}{5}-1}\)
2. \( f(a + b) \le d(a) f(b) + d(b) f(a) \)
Here, \( d(n) \) denotes the number of divisors of \( n... | [
"We need find injective functions satisfying growth bound and inequality. Likely only f(n)=n works? Check condition1: f(n)/n ≤ ln(n)^{d(n)/5 -1}. For f=n, left=1 ≤ ln(n)^{d/5 -1}. For large n, d(n) grows slowly, but d/5 -1 could be negative for small d. For large n, d(n)≥2, so exponent≥ -? For large n, d(n) grows r... | \[
\boxed{\text{There is no injective function }f:\mathbb N\to\mathbb N\text{ that satisfies the two given conditions.}}
\] |
nemotron-math-v2-low-00001058-3bf357c7-03bd-4f21-74e0-72b0c4797ac3 | nvidia/Nemotron-Math-v2/aops | Let \( a \), \( b \), and \( c \) be arbitrary real numbers from the unit-interval \([0,1]\). Determine the inequality bounds and the cases of equality for the expression:
\[
\frac{2+\sqrt{3}}{4} \leq \frac{(ab+b+1)(bc+c+1)(ca+a+1)}{(ab+a+1)(bc+b+1)(ca+c+1)} \leq 8-4\sqrt{3}
\] | [
"We need find min and max of given ratio. Probably symmetry suggests min = (2+√3)/4 and max = 8-4√3? Actually note 8-4√3 ≈ 8-6.928=1.072. The lower bound (2+√3)/4 ≈ (2+1.732)/4=3.732/4=0.933. So ratio between ~0.933 and ~1.072. Likely min when variables all 0 or 1? Test a=b=c=0 => numerator (0+0+1)^3 =1, denominato... | \[
\boxed{\displaystyle
\frac{2+\sqrt3}{4}\;\le\;
\frac{(ab+b+1)(bc+c+1)(ca+a+1)}
{(ab+a+1)(bc+b+1)(ca+c+1)}
\;\le\;8-4\sqrt3 }
\]
The two extreme values are attained only at the following
triples \((a,b,c)\in[0,1]^3\).
* **Lower bound** \(\displaystyle \frac{2+\sqrt3}{4}\)
Equality holds when one of the v... |
nemotron-math-v2-low-00001069-d60f736a-ad26-fcc1-75e9-5cce5407e552 | nvidia/Nemotron-Math-v2/aops | An $n \times n$ grid has the first column and first row completely filled with the integers from $1$ to $n$, with one integer in each box. In how many ways can the remainder of the grid be filled with the integers from $1$ to $n$ so that each row and each column contains each integer exactly once? | [
"We need number of Latin squares of order n with first row and column fixed to 1..n in order. That's known: number = (n-1)! times number of reduced Latin squares? Actually a reduced Latin square has first row and column in natural order. Number of reduced Latin squares is L_n^{red}. The total number of Latin square... | The condition that the first row and the first column are already fixed to
\[
(1,2,\dots ,n)
\]
means that the whole \(n\times n\) array must be a **reduced Latin square** of order \(n\):
a Latin square in which the first row and the first column are in the natural order.
If we denote by
\[
R_n=\text{(number of... |
nemotron-math-v2-low-00001096-0a036684-d215-89a7-7876-544a922a29a6 | nvidia/Nemotron-Math-v2/aops | Find all real numbers \( k \) such that the inequality
\[
\sum_{\text{cyc}} \frac{a}{1 + k(b^2 + c^2)} \geq \frac{3}{1 + 2k}
\]
holds for all positive real numbers \( a, b, c \) with the condition \( a + b + c = 3 \). | [
"We need find k so that inequality holds. Likely using Jensen or Cauchy. Try symmetry, assume a=b=c=1 gives LHS = sum 1/(1+ k(1+1)) =3/(1+2k). RHS same. So equality at a=b=c. Need LHS >= RHS for all. Test extremes: let a→3, b,c→0+. Then term a/(1+k(b^2+c^2)) ≈3/(1+0)=3. Other terms small: b/(1+k(c^2+a^2)) ~0/(1+k*9... | \[
\boxed{0}
\] |
nemotron-math-v2-low-00001102-88a49fd6-3cf5-6026-5bb1-ab8aa92956ce | nvidia/Nemotron-Math-v2/aops | In a table with $n$ rows and $m$ columns, each cell contains one of the letters $a$, $b$, $c$. Every two rows have the same letter in at most $k \geq 0$ positions, and every two columns coincide in at most $k$ positions. Find $m$, $n$, and $k$ if
\[
\frac{2mn + 6k}{3(m + n)} \geq k + 1.
\] | [
"We need find integers m,n,k satisfying condition and inequality? Probably find all? Solve inequality: (2mn+6k)/(3(m+n)) >= k+1. Multiply: 2mn+6k >=3(k+1)(m+n) =3(k+1)m+3(k+1)n. Rearr: 2mn -3(k+1)(m+n)+6k >=0. Consider treat as quadratic in m: 2n m -3(k+1)m -3(k+1)n +6k >=0 => m(2n-3(k+1)) >=3(k+1)n -6k. For positi... | The only possible choice of parameters is
\[
\boxed{m=3,\; n=3,\; k=0}.
\] |
nemotron-math-v2-low-00001104-4039cf49-8f80-eaf7-b89f-18123a8d6707 | nvidia/Nemotron-Math-v2/aops | There are 64 booths around a circular table, each with a chip numbered from 1 to 64. Every minute, chip \(i\) moves \(i\) booths clockwise. At any minute, a light bulb is lit for each chip sharing a booth with chip 1. Determine the position of chip 1 on the first minute when all 1996 light bulbs are lit. | [
"We need time when chip1 shares booth with all other chips? Actually 1996 bulbs meaning total chips? 64 chips only. Maybe bulbs per booth? Actually each chip sharing a booth with chip1 lights a bulb, total bulbs = number of chips that are in same booth as chip1 at that minute. Want 1996 bulbs lit, meaning 1996 chip... | \[
\boxed{32}
\] |
nemotron-math-v2-low-00001120-273c2ad2-c46f-9e51-4342-235ff10d1b20 | nvidia/Nemotron-Math-v2/aops | Find real numbers \(a\), \(b\), and \(c\) such that \((abc)^2 \geq (a+b+c)^3\). | [
"We need find real numbers satisfying inequality. Many solutions exist. Probably they want condition? Maybe find all? We can use AM-GM? Need (abc)^2 >= (a+b+c)^3. If all nonnegative, take a=b=c=t. Then (t^3)^2 = t^6 >= (3t)^3 =27 t^3 => t^6 -27 t^3 = t^3(t^3-27)>=0 => t^3 >=27 => t >=3. So a=b=c>=3 works. Also nega... | To look for real numbers \(a,\;b,\;c\) that satisfy
\[
(abc)^{2}\;\ge\;(a+b+c)^{3},
\]
it is convenient to start with the simplest symmetric choice, namely
\[
a=b=c=t\qquad(t\in\mathbb {R}).
\]
---
### 1. Write the inequality for \(a=b=c=t\)
\[
(abc)^{2}= (t\cdot t\cdot t)^{2}=t^{6},
\qquad
(a+b+c)^{3}= (t+... |
nemotron-math-v2-low-00001129-55f3de51-0d4b-19f6-fe88-258e62f59b43 | nvidia/Nemotron-Math-v2/aops | Let \( f(0) = 0 \), \( f(1) = 1 \), and \( f(n+1) = f(n) + f(n-1) \) for \( n \geq 1 \). Solve the equation in positive integers: \( f(2n+1) = k^2 \). | [
"We need Fibonacci numbers: f is Fibonacci with f(0)=0,f(1)=1 => f(n)=F_n. Condition: F_{2n+1}=k^2 a perfect square. Find all positive integer solutions (n,k). Known that only square Fibonacci numbers are F_0=0, F_1=1, F_12=144=12^2. Also F_{24}=46368 not square. It's known that the only perfect squares in Fibonacc... | The recurrence
\[
f(0)=0,\qquad f(1)=1,\qquad f(n+1)=f(n)+f(n-1)\;(n\ge 1)
\]
produces the Fibonacci numbers:
\[
f(n)=F_n\;(0,1,1,2,3,5,8,13,21,34,\dots )
\]
Hence the equation to be solved is
\[
F_{2n+1}=k^{2},
\qquad n,k\in\mathbb Z_{>0}.
\]
------------------------------------------------------------------... |
nemotron-math-v2-low-00001138-fc2e5bc8-9ab1-09a8-833a-02dabea3f32d | nvidia/Nemotron-Math-v2/aops | For natural numbers $a > b > 1$, define a sequence $x_1, x_2, \ldots$ with the formula $x_n = \frac{a^n - 1}{b^n - 1}$. Find the smallest $d$ such that this sequence has no $d$ consecutive terms that are all prime, regardless of the values of $a$ and $b$. | [
"We need smallest d so that any such sequence cannot have d consecutive primes. For any a>b>1. We need guarantee that among any d consecutive n, some term composite. Find maximal possible length of prime run. Likely small. Try n=1 gives (a-1)/(b-1). Could be prime. For n multiples? Note x_n = (a^n-1)/(b^n-1). Note ... | The longest possible run of consecutive prime values of
\[
x_n=\frac{a^{\,n}-1}{b^{\,n}-1}\qquad (a>b>1)
\]
is two terms.
Indeed, if \(x_k\) and \(x_{k+1}\) were both prime then, using the identities
\[
a^{k+1}-1=(a^{k}-1)a+(a-1),\qquad
b^{k+1}-1=(b^{k}-1)b+(b-1),
\]
one can write
\[
x_{k+1}= \frac{a^{k+1}... |
nemotron-math-v2-low-00001169-94fa3074-33c1-2b44-7688-37a7b8f9c183 | nvidia/Nemotron-Math-v2/aops | Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) such that \( f(1) = 1 \) and
\[ f(f(x)y + \frac{x}{y}) = xyf(x^2 + y^2). \] | [
"We need find solutions. Try assume f(x)=x? Test: LHS f(f(x)y + x/y)= f(x y + x/y). RHS = x y * (x^2 + y^2). If f(t)=t then LHS = xy + x/y. RHS = xy(x^2+y^2). Not equal generally. Try f(x)=k/x? maybe.",
"Let’s try form f(x)=c/x. Then f(1)=c=1 so f(x)=1/x. Test: LHS f(f(x) y + x/y)= f((1/x) y + x/y)= f( y/x + x/y ... | \[
\boxed{\,f(x)=\begin{cases}
\dfrac{1}{x}, & x\neq 0,\\[4pt]
0, & x=0 .
\end{cases}\,}
\]
**Verification**
1. \(f(1)=1\) is satisfied.
2. Let \(y\neq0\).
For \(x\neq0\),
\[
\begin{aligned}
\text{LHS} &= f\!\Bigl(f(x)y+\frac{x}{y}\Bigr)
= f\!\Bigl(\frac{y}{x}+\frac{x}{y}\Bigr) \\
&= ... |
nemotron-math-v2-low-00001185-58a1a8d7-7c96-1c11-bc74-54e9e57b5d18 | nvidia/Nemotron-Math-v2/aops | Solve for \( n \) in the congruence \( n + 11 \equiv 0 \pmod{n-1} \). | [
"We need n such that n+11 divisible by n-1. Let k = n-1. Then n = k+1. Condition: (k+1)+11 ≡0 mod k => k+12 ≡0 mod k => k divides k+12 => k divides 12. So k is a divisor of 12 (positive or maybe allow negative?). Since modulus positive, k>0 => k ∈ divisors of 12: 1,2,3,4,6,12. Then n = k+1 ⇒ n ∈ {2,3,4,5,7,13}. Als... | We need integers \(n\) such that
\[
n+11 \equiv 0 \pmod{n-1},
\]
i.e. \(n-1\) divides \(n+11\).
---
### 1. Introduce a new variable
Let
\[
k = n-1 \qquad\Longrightarrow\qquad n = k+1 .
\]
The condition becomes
\[
k \mid (k+1)+11 = k+12 .
\]
---
### 2. Reduce the divisibility condition
Since \(k\) a... |
nemotron-math-v2-low-00001190-69e8dbdc-7ef1-9225-8186-3315c80d3c01 | nvidia/Nemotron-Math-v2/aops | How many integers are present in the factorial of 3001 (3001!)? | [
"We need number of digits? \"How many integers are present in the factorial of 3001 (3001!)?\" Likely number of digits in 3001!. So compute floor(log10(3001!))+1 using Stirling or sum logs. Approx using Kamenetsky formula. n=3001.",
"Digits = floor( n*log10(n/e) + log10(2πn)/2 ) +1.",
"Compute: log10(e)=0.43429... | To find how many decimal digits are in \(3001!\) we can use the well‑known formula for the number of digits of \(n!\):
\[
\text{digits}(n!)=\left\lfloor n\log_{10}\!\left(\frac{n}{e}\right)+\frac{1}{2}\log_{10}(2\pi n)\right\rfloor+1 .
\]
---
### Apply the formula for \(n=3001\)
1. **Compute \( \log_{10}\!\left(\fr... |
nemotron-math-v2-low-00001225-cab64f77-4de6-c56d-61c3-c948d216b268 | nvidia/Nemotron-Math-v2/aops | Let \( c \) be a composite number. What is the limit of \( A \)?
\[ A = \lim_{n \to \infty} \sum_{1 \leq c \leq n} \mu(c) \] | [
"We need sum of Möbius up to n over composite c only. Sum over all c≤n μ(c) = M(n) partial sums of Möbius. Known that sum over all integers up to n of μ(k) = o(n) and actually tends to 0? Mertens conjecture false? Actually M(n)=o(n) and limit as n→∞ M(n)/n =0. So sum grows slower than n, but does it converge? It di... | The sum in \(A\) runs over the composite integers only.
Recall the Möbius function
\[
\mu(k)=\begin{cases}
1 & k=1,\\[2pt]
(-1)^{r} & k \text{ is square‑free with } r \text{ distinct prime factors},\\[2pt]
0 & \text{otherwise}.
\end{cases}
\]
Hence for a composite integer \(c\),
* if \(c\) is not square‑free, \(... |
nemotron-math-v2-low-00001233-dc53d5ec-5555-53e6-26d6-23185514eab9 | nvidia/Nemotron-Math-v2/aops | Find all functions \( f: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+} \) that satisfy \( f(f(x) + y) \mid x + f(y) \) for all \( x, y \in \mathbb{Z}^{+} \). | [
"We need f: positive integers to positive integers. Condition: f(f(x)+y) divides x+f(y). Try linear f(x)=cx. Suppose f(x)=kx. Then LHS f(kx + y)=k(kx + y)=k^2 x + k y divides x + k y. So k^2 x + k y ≤ x + k y => (k^2 -1)x ≤0 for all x>0, so k^2-1 ≤0 => k=1. k=1 works? Check: f(x)=x. Then f(f(x)+y)=f(x+y)=x+y divide... | \[
\boxed{\;f(x)=1\ \text{for all }x\in\mathbb Z^{+}\quad\text{or}\quad f(x)=x\ \text{for all }x\in\mathbb Z^{+}\;}
\] |
nemotron-math-v2-low-00001257-cd04963a-af86-a83e-50c6-67562d8c3172 | nvidia/Nemotron-Math-v2/aops | Find all perfect squares \( x \) such that \( x \) contains every digit (0, 1, ..., 9) exactly once. | [
"We need 10-digit pandigital perfect squares. Search constraints: 10-digit number uses each digit once. Must be a perfect square. Range: smallest 10-digit number = 1,023,456,789? Actually smallest 10-digit = 1,023,456,789? No 10-digit minimum is 1,023,456,789? That's 9 digits. 10-digit minimum is 1,023,456,789? Wai... | A 10‑digit number that uses each of the digits \(0,1,\dots ,9\) exactly once must lie between
\[
10^{9}=1\,000\,000\,000 \qquad\text{and}\qquad 9\,999\,999\,999 .
\]
Hence its square root lies between
\[
\sqrt{10^{9}}\approx 31\,623 \quad\text{and}\quad \sqrt{10^{10}}\approx 99\,500 .
\]
Scanning the squares of... |
nemotron-math-v2-low-00001276-d9bb5ace-2fe2-538b-e59b-9008664160a7 | nvidia/Nemotron-Math-v2/aops | What is the largest number of real roots that a polynomial \( p(x) \) with real coefficients can have, given that \( p(x^2 - 1) = p(x) p(-x) \)? | [
"We need constraints. Let p be polynomial real coeff satisfying functional equation. Let degree n. Then deg LHS = n * deg(x^2-1) = 2n. RHS deg = n + n = 2n. Leading coefficient: Let a be leading coeff of p. Then LHS leading term a*(x^2)^n = a x^{2n}. RHS leading term a x^n * a (-x)^n = a^2 * (-1)^n x^{2n}. So a = a... | \[
\boxed{4}
\] |
nemotron-math-v2-low-00001286-a6861d0b-d1cf-18ec-b9cb-8b49f3413b33 | nvidia/Nemotron-Math-v2/aops | Given the sequence $(a_n)$ defined by $a_0 = 0$, $a_1 = 1$, and $a_{n+1} = 2a_n + (a-1)a_{n-1}$ for all $n \ge 1$, where $a \in \mathbb{N}$, and an odd prime number $q$, find the smallest value of $a$ such that:
$$\begin{cases}
p \mid a_p \ \ \forall p \le q \\
p \nmid a_p \ \ \forall p > q
\end{cases}$$
where $p$ is a... | [
"We need a such that a_p ≡0 mod p for all primes ≤ q, and not divisible for primes > q. Likely a = q? Test small. Compute recurrence modulo p. Characteristic x^2-2x-(a-1)=0 => roots r=1±√a. Then a_n = ((1+√a)^n - (1-√a)^n)/(2√a). So a_p mod p using Fermat? For prime p, if √a exists mod p then expression ≡0? Need a ... | **Answer**
\[
\boxed{a = \displaystyle \prod_{p\le q}p}
\] |
nemotron-math-v2-low-00001312-37b75678-f996-5bb8-2c52-4162c6fd967a | nvidia/Nemotron-Math-v2/aops | Determine the maximum number of points in the interior of an m×n grid (including the boundary), such that no two points are at a distance less than $\sqrt{2}$. | [
"We need max points on integer lattice? interior of grid including boundary means points with integer coordinates? Likely we consider points at grid points (including boundaries) in an m×n rectangular lattice. Need no two points distance < sqrt2. Minimum distance sqrt2 corresponds to adjacent orthogonal neighbors d... | The condition “no two points are at a distance less than \(\sqrt{2}\)” means that two points may **not** be at Euclidean distance \(1\).
On a rectangular grid the only pairs of points at distance \(1\) are those that lie in adjacent lattice squares (horizontally or vertically).
Hence we must choose a set of lattic... |
nemotron-math-v2-low-00001324-08d5fb24-7d4d-7b00-72a7-243a7395058e | nvidia/Nemotron-Math-v2/aops | In triangle \(ABC\), \(AB = 10\) and \(CA = 12\). The angle bisector of \(\angle BAC\) intersects \(CA\) at \(E\), and the angle bisector of \(\angle ABC\) intersects \(AB\) at \(D\). \(AM\) and \(AN\) are the perpendiculars from \(A\) to \(CD\) and \(BE\) respectively. If \(MN = 4\), find the length of \(BC\). | [
"We need relation. Likely using property that distance between feet of perpendiculars from vertex to angle bisector lines equals something involving side lengths. Might be formula: MN = (bc/ (b+c)) * sin(A/2)?? Not sure.",
"Let sides: AB=c=10, AC=b=12, BC=a = ?. Angles at A opposite a. Angle bisector from A meets... | \[
\boxed{14}
\] |
nemotron-math-v2-low-00001337-8f2dd929-97d4-afa9-08e2-1a82a95bf261 | nvidia/Nemotron-Math-v2/aops | Given that the ratio of the area of a triangle to the area of its contact triangle is \(8:3\), what is the ratio of the circumradius to the inradius of the triangle? | [
"We need relation. For triangle, area ratio = (R/r)*(s/R?) Actually contact triangle (intouch) area = r*s/2? Wait original area = Δ = rs. Contact triangle area = Δ_intouch = r^2 * s / (2R)? Not sure. Known formula: area of intouch triangle = r * s * (R-2r)/4R? Hmm.",
"Better: sides a,b,c. Contact triangle vertice... | For a triangle \(ABC\) let
* \(R\) be its circumradius,
* \(r\) be its inradius,
* \(\Delta\) be its area, and
* \(\Delta_{I}\) be the area of the contact (intouch) triangle formed by the three points where the incircle touches the sides.
A classical formula for the area of the intouch triangle is
\[
\Delt... |
nemotron-math-v2-low-00001343-c4bf376f-e076-b246-09a2-a252d2667cb6 | nvidia/Nemotron-Math-v2/aops | Solve the system of equations:
\[ x^n - 6y^2 + 11z = 3, \]
\[ y^n - 6z^2 + 11x = 3, \]
\[ z^n - 6x^2 + 11y = 3. \] | [
"We need solve for real numbers? Likely integer? n is given? Probably n is a positive integer. Symmetry suggests solutions where x=y=z. Let x=y=z=a. Then each equation: a^n -6a^2 +11a =3. So a satisfies f(a)=a^n -6a^2 +11a -3=0.",
"We need find a for any n? Try small n values. For n=1: a -6a^2 +11a -3 = -6a^2 +12... | Because the three equations are obtained from one another by a cyclic permutation of the variables, any solution must satisfy
\[
x=y=z.
\]
(If two of the variables were different we could subtract two of the equations and obtain a contradiction; the detailed verification is omitted here because the cyclic symmetry ... |
nemotron-math-v2-low-00001361-7f21286f-0cd8-df79-85ab-e56fc7c52974 | nvidia/Nemotron-Math-v2/aops | Find all functions \( f : \mathbb{R} \to \mathbb{R} \) that satisfy the equation:
\[ f(xy + y) + f(x) = f(xy + x) + f(y) \] | [
"We need Cauchy type. Let’s set y=0: f(0+0)+f(x)=f(0+x)+f(0) => f(0)+f(x)=f(x)+f(0) holds trivially.",
"Let’s set x=1: f(y + y)+f(1)=f(y +1)+f(y) => f(2y)+f(1)=f(y+1)+f(y). (1)",
"Similarly y=1: f(x+1)+f(x)=f(x + x)+f(1) => f(x+1)+f(x)=f(2x)+f(1). (2)",
"From (1) and (2) we have symmetry: f(2y)+f(1)=f(y+1)+f(... | \[
\boxed{\;f(x)=ax+b\qquad (a,b\in\mathbb{R})\;}
\] |
nemotron-math-v2-low-00001375-b1d1d971-3ad7-c786-55a8-6b6458999b95 | nvidia/Nemotron-Math-v2/aops | Take a triangle. Trisect its sides and cut off the small triangle formed by a vertex of the triangle and two of the trisection points in each corner to get a hexagon. Repeat this operation on the hexagon, trisecting the sides and chopping the corners to get a dodecagon. Repeat this process ad infinitum. What does the r... | [
"We need description: the limit shape is a fractal curve? Actually process: each iteration removes three small similar triangles from each corner, leaving a polygon with twice the sides each time. In limit, becomes a curve known as the Koch snowflake? Not exactly. Scaling factor? At each step, side length multiplie... | \[
\boxed{\text{The limiting figure is a convex, infinitely‑sided curve – a genuine fractal –
but it is **not** a conic (ellipse, parabola, hyperbola, etc.).}}
\] |
nemotron-math-v2-low-00001394-e2fd0056-6c14-26d0-d400-22969c4eccfa | nvidia/Nemotron-Math-v2/aops | Find integers \(a\), \(b\), and \(c\) such that \((a + b + c)^2 = abc\). | [
"We need integer solutions to (a+b+c)^2 = a b c. Look for nontrivial. Try small values. If any zero? If any of a,b,c =0 then RHS=0 => (sum)^2 =0 => sum=0. So any triple with one zero and sum zero works. e.g., a=0, b=1, c=-1. That's integer solution. Also all zero works. If none zero, we can divide. Suppose all nonz... | \[
\boxed{a=9,\;b=9,\;c=9}
\]
Indeed
\[
(a+b+c)^2=(9+9+9)^2=27^2=729,
\qquad
abc=9\cdot9\cdot9=729.
\]
Hence \((a+b+c)^2=abc\).
(Other integer solutions exist – for example any triple with one entry equal to \(0\) and the other two opposite numbers, such as \((0,\,1,\,-1)\) – but the triple \((9,9,9)\) is a non‑... |
nemotron-math-v2-low-00001413-f6f50cae-d31c-e36f-705c-8a84039726e6 | nvidia/Nemotron-Math-v2/aops | Find all solutions to the Diophantine equation \(1 + x + x^2 + \ldots + x^a = y^b\) where \((x, y, a, b) \in \mathbb{Z}^4\) are integers with \(\min\{a, b, |x|\} > 1\) in the case where \(4 \mid a + 1\). | [
"We need solutions to sum geometric series = (x^{a+1}-1)/(x-1)= y^b. with a+1 divisible by 4. Known results: Catalan? Probably only trivial: x=3,a=3 gives 1+3+9+27=40 not perfect power. Need known theorem: Nagell–Ljunggren equation (x^{n}-1)/(x-1)=y^q. Solutions known: (x,n,y,q) = (3,5,11,2) gives 1+3+9+27+81=121=1... | The equation
\[
1+x+x^{2}+\cdots +x^{a}=y^{b}, \qquad (x,y,a,b)\in\mathbb Z^{4},
\qquad
\min\{a,b,|x|\}>1,
\]
can be rewritten as
\[
\frac{x^{a+1}-1}{x-1}=y^{b}. \tag{1}
\]
Equation (1) is the classical **Nagell–Ljunggren equation**.
All integer solutions of (1) with \(|x|>1,\; a+1>1\) are known (see for exa... |
nemotron-math-v2-low-00001433-3e5ae993-7cf1-6f2e-c195-1c3e4e142e36 | nvidia/Nemotron-Math-v2/aops | Determine the least value of the number of a sequence such that there exists a way to reorder it so that the sum of any three adjacent numbers is divisible by the previous number, and the last number in the sequence is odd. | [
"We need interpret: sequence of numbers? Probably positive integers 1..n? Asking least value of the number of a sequence (i.e., length) such that there exists a permutation where for each position i (i>=2?), condition: sum of any three adjacent numbers is divisible by the previous number. \"previous number\" meanin... | We look for the smallest length \(n\) of a sequence (using the numbers \(1,2,\dots ,n\)) that can be permuted so that
* for every block of three consecutive terms \(a_i ,a_{i+1},a_{i+2}\) the sum
\(a_i+a_{i+1}+a_{i+2}\) is divisible by the term that comes **just before** this block,
i.e. by \(a_{i-1}\);
* the la... |
End of preview. Expand in Data Studio
Nemotron Math V2 Low Mini
A compact textual reasoning dataset derived from the low 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
low-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 | 1,718,159 |
| Final records | 58,595 |
| Retention | 3.410% |
| File size (canonical JSONL) | 145,436,648 bytes |
| Sources represented | 2 |
| Total reasoning steps | 328,464 |
| Mean steps per record | 5.61 |
| Median steps per record | 5 |
| Steps per record | 3-46 |
| Mean individual step length | 179.68 characters |
| Median individual step length | 103 characters |
| Individual step-length range | 2-1,024 characters |
| Median total reasoning length | 902 characters |
| Total reasoning-length range | 129-3,698 characters |
| Answer-length range | 9-3,143 characters |
| Question-length range | 14-1,102 characters |
Source composition
| Source | Records |
|---|---|
nvidia/Nemotron-Math-v2/stackflow |
38,857 |
nvidia/Nemotron-Math-v2/aops |
19,738 |
| Total | 58,595 |
Rejection accounting
| Reason | Rows |
|---|---|
| Source conversation is not exactly two turns | 462,352 |
| Tool trajectory | 412,640 |
| Fewer than 3 reasoning steps | 352,792 |
| Complete Llama 3 sequence over 4,096 characters | 180,422 |
| Complete Llama 3 sequence over 1,024 tokens | 134,549 |
| A reasoning step over 1,024 characters | 106,411 |
| More than 50 reasoning steps | 10,290 |
| User message does not end with exact problem | 103 |
| Fewer than 128 reasoning characters | 5 |
| Accepted | 58,595 |
| Total | 1,718,159 |
Integrity and provenance
The canonical local JSONL has SHA-256 5bc93ca130c45f1b94df993cc5f55127d2431b4010655ec6b2ee40dc4a39189a. 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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