Datasets:
task_id stringlengths 26 34 | input stringlengths 189 1.32k | raw_output stringlengths 327 135k | thinking_content null | final_answer stringlengths 327 135k | finish_reason stringclasses 2
values | prompt_tokens int64 60 636 | completion_tokens int64 110 32.8k | temperature float64 0 0 | max_tokens int64 32.8k 32.8k |
|---|---|---|---|---|---|---|---|---|---|
AMC_12__2012_AMC_12A_Problems__p21 | Let $a$, $b$, and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that
$a^2-b^2-c^2+ab=2011$ and
$a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$.
What is $a$?
$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$
Please reason step by step, and put your final answer wit... | Let the given equations be
$$a^2-b^2-c^2+ab=2011 \quad (*)$$
$$a^2+3b^2+3c^2-3ab-2ac-2bc=-1997 \quad (**)$$
We can rewrite the second equation as
$$a^2+3b^2+3c^2-3ab-2ac-2bc = a^2-ab-ac + 2b^2+2c^2-2ab-ac-2bc = -1997$$
We can rewrite the second equation as
$$a^2+3b^2+3c^2-3ab-2ac-2bc = (a-b-c)^2 + 2b^2+2c^2+2bc-2ab-2ac... | null | Let the given equations be
$$a^2-b^2-c^2+ab=2011 \quad (*)$$
$$a^2+3b^2+3c^2-3ab-2ac-2bc=-1997 \quad (**)$$
We can rewrite the second equation as
$$a^2+3b^2+3c^2-3ab-2ac-2bc = a^2-ab-ac + 2b^2+2c^2-2ab-ac-2bc = -1997$$
We can rewrite the second equation as
$$a^2+3b^2+3c^2-3ab-2ac-2bc = (a-b-c)^2 + 2b^2+2c^2+2bc-2ab-2ac... | length | 172 | 32,768 | 0 | 32,768 |
AMC_10__2014_AMC_10A_Problems__p3 | Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $\textdollar 2.50$ each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf co... | Let $N$ be the number of loaves of bread Bridget bakes. We are given that $N = 48$.
In the morning, she sells half of the loaves, which is $\frac{1}{2} \times 48 = 24$ loaves. She sells them for $\textdollar 2.50$ each, so she earns $24 \times 2.50 = \textdollar 60$.
The number of loaves remaining after the morning is ... | null | Let $N$ be the number of loaves of bread Bridget bakes. We are given that $N = 48$.
In the morning, she sells half of the loaves, which is $\frac{1}{2} \times 48 = 24$ loaves. She sells them for $\textdollar 2.50$ each, so she earns $24 \times 2.50 = \textdollar 60$.
The number of loaves remaining after the morning is ... | stop | 181 | 396 | 0 | 32,768 |
AMC_12__2018_AMC_12A_Problems__p6 | For positive integers $m$ and $n$ such that $m+10<n+1$, both the mean and the median of the set $\{m, m+4, m+10, n+1, n+2, 2n\}$ are equal to $n$. What is $m+n$?
$\textbf{(A)}20\qquad\textbf{(B)}21\qquad\textbf{(C)}22\qquad\textbf{(D)}23\qquad\textbf{(E)}24$
Please reason step by step, and put your final answer within... | Let the set be $S = \{m, m+4, m+10, n+1, n+2, 2n\}$.
The mean of the set is $\frac{m + (m+4) + (m+10) + (n+1) + (n+2) + 2n}{6} = \frac{3m + 4n + 17}{6}$.
We are given that the mean is $n$, so $\frac{3m + 4n + 17}{6} = n$.
Multiplying by 6, we get $3m + 4n + 17 = 6n$, which simplifies to $3m - 2n = -17$.
We are also gi... | null | Let the set be $S = \{m, m+4, m+10, n+1, n+2, 2n\}$.
The mean of the set is $\frac{m + (m+4) + (m+10) + (n+1) + (n+2) + 2n}{6} = \frac{3m + 4n + 17}{6}$.
We are given that the mean is $n$, so $\frac{3m + 4n + 17}{6} = n$.
Multiplying by 6, we get $3m + 4n + 17 = 6n$, which simplifies to $3m - 2n = -17$.
We are also gi... | stop | 142 | 812 | 0 | 32,768 |
USAMO__2019_USAMO_Problems__p4 | Let $n$ be a nonnegative integer. Determine the number of ways that one can choose $(n+1)^2$ sets $S_{i,j}\subseteq\{1,2,\ldots,2n\}$, for integers $i,j$ with $0\leq i,j\leq n$, such that:
$\bullet$ for all $0\leq i,j\leq n$, the set $S_{i,j}$ has $i+j$ elements; and
$\bullet$ $S_{i,j}\subseteq S_{k,l}$ whenever $0\leq... | Let $n$ be a nonnegative integer. We want to determine the number of ways to choose $(n+1)^2$ sets $S_{i,j} \subseteq \{1, 2, \dots, 2n\}$ for integers $i, j$ with $0 \le i, j \le n$, such that:
\begin{itemize}
\item for all $0 \le i, j \le n$, the set $S_{i,j}$ has $i+j$ elements; and
\item $S_{i,j} \subseteq ... | null | Let $n$ be a nonnegative integer. We want to determine the number of ways to choose $(n+1)^2$ sets $S_{i,j} \subseteq \{1, 2, \dots, 2n\}$ for integers $i, j$ with $0 \le i, j \le n$, such that:
\begin{itemize}
\item for all $0 \le i, j \le n$, the set $S_{i,j}$ has $i+j$ elements; and
\item $S_{i,j} \subseteq ... | stop | 174 | 2,066 | 0 | 32,768 |
AMC_8__1996_AJHSME_Problems__p25 | A point is chosen at random from within a circular region. What is the probability that the point is closer to the center of the region than it is to the boundary of the region?
$\text{(A)}\frac{1}{4} \qquad \text{(B)}\frac{1}{3} \qquad \text{(C)}\frac{1}{2} \qquad \text{(D)}\frac{2}{3} \qquad \text{(E)}\frac{3}{4}$
... | Let the radius of the circular region be $R$. Let $(x, y)$ be a point chosen at random from within the circular region. The distance from the point $(x, y)$ to the center of the region is $\sqrt{x^2 + y^2}$. The distance from the point $(x, y)$ to the boundary of the region is $R - \sqrt{x^2 + y^2}$.
We want to find th... | null | Let the radius of the circular region be $R$. Let $(x, y)$ be a point chosen at random from within the circular region. The distance from the point $(x, y)$ to the center of the region is $\sqrt{x^2 + y^2}$. The distance from the point $(x, y)$ to the boundary of the region is $R - \sqrt{x^2 + y^2}$.
We want to find th... | stop | 131 | 422 | 0 | 32,768 |
AIME__2004_AIME_II_Problems__p3 | A solid rectangular block is formed by gluing together $N$ congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly $231$ of the 1-cm cubes cannot be seen. Find the smallest possible value of $N.$
Please reason step by step, and put your final answer within oxed{}. | Let the dimensions of the rectangular block be $x, y, z$, where $x, y, z$ are positive integers. The total number of cubes is $N = xyz$.
When the block is viewed so that three of its faces are visible, the number of cubes that cannot be seen is $(x-1)(y-1)(z-1)$.
We are given that $(x-1)(y-1)(z-1) = 231$. We want to fi... | null | Let the dimensions of the rectangular block be $x, y, z$, where $x, y, z$ are positive integers. The total number of cubes is $N = xyz$.
When the block is viewed so that three of its faces are visible, the number of cubes that cannot be seen is $(x-1)(y-1)(z-1)$.
We are given that $(x-1)(y-1)(z-1) = 231$. We want to fi... | stop | 91 | 1,117 | 0 | 32,768 |
AHSME__1979_AHSME_Problems__p8 | Find the area of the smallest region bounded by the graphs of $y=|x|$ and $x^2+y^2=4$.
$\textbf{(A) }\frac{\pi}{4}\qquad \textbf{(B) }\frac{3\pi}{4}\qquad \textbf{(C) }\pi\qquad \textbf{(D) }\frac{3\pi}{2}\qquad \textbf{(E) }2\pi$
Please reason step by step, and put your final answer within oxed{}. | Let the region be $R$. The graphs of $y=|x|$ and $x^2+y^2=4$ are a V-shaped graph and a circle with radius 2 centered at the origin, respectively.
We need to find the points of intersection of the two graphs.
For $x \ge 0$, we have $y=x$, so $x^2+x^2=4$, which gives $2x^2=4$, so $x^2=2$, and $x=\sqrt{2}$. Then $y=\sqrt... | null | Let the region be $R$. The graphs of $y=|x|$ and $x^2+y^2=4$ are a V-shaped graph and a circle with radius 2 centered at the origin, respectively.
We need to find the points of intersection of the two graphs.
For $x \ge 0$, we have $y=x$, so $x^2+x^2=4$, which gives $2x^2=4$, so $x^2=2$, and $x=\sqrt{2}$. Then $y=\sqrt... | stop | 121 | 936 | 0 | 32,768 |
AMC_10__2002_AMC_10A_Problems__p2 | Given that a, b, and c are non-zero real numbers, define $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$, find $(2, 12, 9)$.
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$
Please reason step by step, and put your final answer within oxed{}. | We are given that $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$. We want to find $(2, 12, 9)$.
We have $a = 2$, $b = 12$, and $c = 9$.
Then
$$(2, 12, 9) = \frac{2}{12} + \frac{12}{9} + \frac{9}{2} = \frac{1}{6} + \frac{4}{3} + \frac{9}{2}$$
To add these fractions, we need a common denominator. The least common ... | null | We are given that $(a, b, c) = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}$. We want to find $(2, 12, 9)$.
We have $a = 2$, $b = 12$, and $c = 9$.
Then
$$(2, 12, 9) = \frac{2}{12} + \frac{12}{9} + \frac{9}{2} = \frac{1}{6} + \frac{4}{3} + \frac{9}{2}$$
To add these fractions, we need a common denominator. The least common ... | stop | 132 | 294 | 0 | 32,768 |
AMC_10__2005_AMC_10A_Problems__p11 | A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$?
$\textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7$
Please reason step by s... | Let $n$ be the side length of the wooden cube. The cube is cut into $n^3$ unit cubes.
The total number of faces of the $n^3$ unit cubes is $6n^3$.
The number of red faces is given to be exactly one-fourth of the total number of faces of the unit cubes, so the number of red faces is $\frac{1}{4}(6n^3) = \frac{3}{2}n^3$.... | null | Let $n$ be the side length of the wooden cube. The cube is cut into $n^3$ unit cubes.
The total number of faces of the $n^3$ unit cubes is $6n^3$.
The number of red faces is given to be exactly one-fourth of the total number of faces of the unit cubes, so the number of red faces is $\frac{1}{4}(6n^3) = \frac{3}{2}n^3$.... | stop | 130 | 959 | 0 | 32,768 |
AHSME__1960_AHSME_Problems__p15 | Triangle $I$ is equilateral with side $A$, perimeter $P$, area $K$, and circumradius $R$ (radius of the circumscribed circle).
Triangle $II$ is equilateral with side $a$, perimeter $p$, area $k$, and circumradius $r$. If $A$ is different from $a$, then:
$\textbf{(A)}\ P:p = R:r \text{ } \text{only sometimes} \qquad \... | Let $A$ be the side length of triangle $I$, and $a$ be the side length of triangle $II$.
The perimeter of an equilateral triangle with side length $s$ is $3s$. Thus, $P = 3A$ and $p = 3a$.
The area of an equilateral triangle with side length $s$ is $\frac{\sqrt{3}}{4}s^2$. Thus, $K = \frac{\sqrt{3}}{4}A^2$ and $k = \fr... | null | Let $A$ be the side length of triangle $I$, and $a$ be the side length of triangle $II$.
The perimeter of an equilateral triangle with side length $s$ is $3s$. Thus, $P = 3A$ and $p = 3a$.
The area of an equilateral triangle with side length $s$ is $\frac{\sqrt{3}}{4}s^2$. Thus, $K = \frac{\sqrt{3}}{4}A^2$ and $k = \fr... | stop | 221 | 690 | 0 | 32,768 |
AMC_12__2023_AMC_12B_Problems__p4 | Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?
$\textbf{(A) } 162,500 \qquad\textbf{(B) } 162.5 \qquad\textbf{(C) }1,625 \qquad\textbf{(D) }1,625,000 \qquad\text... | Let the width of the strip be $w = 6.5$ millimeters.
Let the length of the strip be $l = 25$ meters.
We want to find the area of the strip in square centimeters.
First, we need to convert the length and width to the same units. Let's convert everything to centimeters.
Since 1 meter = 100 centimeters, we have $l = 25 \t... | null | Let the width of the strip be $w = 6.5$ millimeters.
Let the length of the strip be $l = 25$ meters.
We want to find the area of the strip in square centimeters.
First, we need to convert the length and width to the same units. Let's convert everything to centimeters.
Since 1 meter = 100 centimeters, we have $l = 25 \t... | stop | 151 | 272 | 0 | 32,768 |
AIME__2015_AIME_I_Problems__p1 | The expressions $A$ = $1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39$ and $B$ = $1 + 2 \times 3 + 4 \times 5 + \cdots + 36 \times 37 + 38 \times 39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference betw... | Let $A = 1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39$.
We can write $A$ as
$$A = \sum_{n=1}^{19} (2n-1)(2n) + 39 = \sum_{n=1}^{19} (4n^2 - 2n) + 39 = 4 \sum_{n=1}^{19} n^2 - 2 \sum_{n=1}^{19} n + 39.$$
We know that $\sum_{n=1}^k n = \frac{k(k+1)}{2}$ and $\sum_{n=1}^k n^2 = \frac{k(k+1)(2k+1)}{6}$... | null | Let $A = 1 \times 2 + 3 \times 4 + 5 \times 6 + \cdots + 37 \times 38 + 39$.
We can write $A$ as
$$A = \sum_{n=1}^{19} (2n-1)(2n) + 39 = \sum_{n=1}^{19} (4n^2 - 2n) + 39 = 4 \sum_{n=1}^{19} n^2 - 2 \sum_{n=1}^{19} n + 39.$$
We know that $\sum_{n=1}^k n = \frac{k(k+1)}{2}$ and $\sum_{n=1}^k n^2 = \frac{k(k+1)(2k+1)}{6}$... | stop | 145 | 1,137 | 0 | 32,768 |
AMC_8__2008_AMC_8_Problems__p24 | Ten tiles numbered $1$ through $10$ are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?
$\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{11}{60}\qquad\textbf{(D)}\... | Let $T$ be the number on the tile, so $T \in \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Let $D$ be the number rolled on the die, so $D \in \{1, 2, 3, 4, 5, 6\}$.
We want to find the probability that $T \cdot D$ is a perfect square.
The total number of possible outcomes is $10 \times 6 = 60$.
We need to find the number of pair... | null | Let $T$ be the number on the tile, so $T \in \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Let $D$ be the number rolled on the die, so $D \in \{1, 2, 3, 4, 5, 6\}$.
We want to find the probability that $T \cdot D$ is a perfect square.
The total number of possible outcomes is $10 \times 6 = 60$.
We need to find the number of pair... | stop | 149 | 837 | 0 | 32,768 |
AMC_12__2020_AMC_12B_Problems__p14 | Bela and Jenn play the following game on the closed interval $[0, n]$ of the real number line, where $n$ is a fixed integer greater than $4$. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval $[0, n]$. Thereafter, the player whose turn it is chooses a real n... | Let $n$ be a fixed integer greater than 4. Bela and Jenn play a game on the interval $[0, n]$. Bela goes first and chooses a number $x_1 \in [0, n]$. Then Jenn chooses a number $x_2$ such that $|x_2 - x_i| > 1$ for all $i=1$. They continue taking turns, choosing numbers that are more than one unit away from all previou... | null | Let $n$ be a fixed integer greater than 4. Bela and Jenn play a game on the interval $[0, n]$. Bela goes first and chooses a number $x_1 \in [0, n]$. Then Jenn chooses a number $x_2$ such that $|x_2 - x_i| > 1$ for all $i=1$. They continue taking turns, choosing numbers that are more than one unit away from all previou... | stop | 242 | 1,465 | 0 | 32,768 |
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google/gemma-3-27b-it — creativemath-with-answers
Model outputs from the micro-creativity inference suite.
- Model:
google/gemma-3-27b-it - Dataset:
creativemath-with-answers(188 items) - Part of collection: ZachW/llm-creativity-benchmarks
Generation config
temperature: 0.0
max_tokens: 32768
seed: 42
backend: vllm
Columns
| Column | Description |
|---|---|
task_id |
Unique task identifier |
input |
The exact prompt sent to the model (after meta-prompt application) |
raw_output |
Full model output string |
thinking_content |
Extracted chain-of-thought / thinking block (null for non-thinking models) |
final_answer |
Extracted final answer after thinking is removed |
finish_reason |
stop (completed) or length (truncated at max_tokens) |
prompt_tokens |
Number of tokens in the prompt |
completion_tokens |
Number of tokens generated |
temperature |
Sampling temperature used |
max_tokens |
Token generation limit |
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