GITHUB HUGGING FACE MODELSCOPE DISCORD

🚨 此模型目前主要支持英语。我们将尽快推出中英双语版本。

# 简介

在过去的一年里，我们非常关注大模型的推理能力的提升，尤其关注其在数学相关的任务上的表现。今天，我们非常高兴地介绍 Qwen2 开源家族的新成员——Qwen2-Math-1.5B/7B/72B 系列。Qwen2-Math 是一系列基于 Qwen2 LLM 构建的专门用于数学解题的语言模型，其数学能力显著超越了开源模型，甚至超过了闭源模型（如 GPT-4o）。我们希望Qwen2-Math能够为科学界解决需要复杂多步逻辑推理的高级数学问题做出贡献。

我们在一系列数学基准评测上评估了我们的数学专用模型 Qwen2-Math。在 Math 上的评测结果表明，我们最大的数学专用模型 Qwen2-Math-72B-Instruct 超越了最先进的模型，包括 GPT-4o、Claude-3.5-Sonnet、Gemini-1.5-Pro 和 Llama-3.1-405B。

## Qwen2-Math基础模型

Qwen2-Math 的基础模型使用 Qwen2-1.5B/7B/72B 进行初始化，然后在精心设计的数学专用语料库上进行预训练，该语料库包含大规模高质量的数学网络文本、书籍、代码、考试题目以及由 Qwen2 模型合成的数学预训练数据。

我们在三个广泛使用的英语数学基准 GSM8K、Math 和 MMLU-STEM 上评估了我们的 Qwen2-Math 基模型。此外，我们还评估了三个中国数学基准 CMATH，GaoKao Math Cloze 和 GaoKao Math QA。所有评估均使用 Few-shot CoT 方式。

## Qwen2-Math指令微调模型

我们首先基于 Qwen2-Math-72B 训练了一个数学专用的奖励模型。然后，我们将这个密集的奖励信号与一个二元信号结合，该二元信号指示模型是否正确回答了问题。这个组合信号被用作监督来通过拒绝采样构建 SFT 数据，并在此SFT模型的基础上进一步使用 GRPO 来优化模型。

我们对 Qwen2-Math-Instruct 在英语和中文的数学基准评测上进行了评估。除了常用的基准评测，如 GSM8K 和 MATH 之外，我们还加入了更具挑战性的考试以全面检测 Qwen2-Math-Instruct 的能力，例如 OlympiadBench、CollegeMath、高考（GaoKao）、AIME2024 以及 AMC2023。对于中文的数学评测集，我们使用了 CMATH、2024年中国高考数学题以及2024年中国中考数学题。

我们汇报了在 greedy 和 RM@8 的条件下模型在各个基准评测上的 0-shot 成绩，对于选择题类型的题目，如 MMLU(STEM)，我们使用了 5-shot 进行评测。

## 案例分析

这里我们列举了一些测试的竞赛题，其中包括了多道 IMO 竞赛题。通过评测及具体样例分析，我们发现 Qwen2-Math 已经具备了解决一些简单竞赛题的数学能力。欢迎点击下方折叠块查看样例！

所有解决方案均由我们的模型生成，未经过任何修改。请注意，我们不保证过程中的陈述是正确的。

## Problem From IMO Shortlist 2002

**Problem**

What is the smallest positive integer $t$ such that there exist integers $x_1,x_2,\ldots,x_t$ with

$$ x^3_1+x^3_2+,\ldots,+x^3_t=2002^{2002}, $$

**Qwen2-Math-72B-Instruct Solution**

To determine the smallest positive integer $ t $ such that there exist integers $ x_1, x_2, \ldots, x_t $ satisfying the equation

$$ x_1^3 + x_2^3 + \cdots + x_t^3 = 2002^{2002}, $$

we start by analyzing the properties of cubes and their sums. We need to find the smallest $ t $ for which the sum of $ t $ cubes equals $ 2002^{2002} $.

First, we note that $ 2002 $ can be factored into prime factors:

$$ 2002 = 2 \times 7 \times 11 \times 13. $$

Thus,

$$ 2002^{2002} = (2 \times 7 \times 11 \times 13)^{2002} = 2^{2002} \times 7^{2002} \times 11^{2002} \times 13^{2002}. $$

Next, we consider the properties of cubes modulo 9. Cubes modulo 9 can only be congruent to $ 0, 1, $ or $ -1 $. This is because:

$$ \begin{aligned} 0^3 &\equiv 0 \pmod{9}, \\\ 1^3 &\equiv 1 \pmod{9}, \\\ 2^3 &\equiv 8 \equiv -1 \pmod{9}, \\\ 3^3 &\equiv 27 \equiv 0 \pmod{9}, \\\ 4^3 &\equiv 64 \equiv 1 \pmod{9}, \\\ 5^3 &\equiv 125 \equiv -1 \pmod{9}, \\\ 6^3 &\equiv 216 \equiv 0 \pmod{9}, \\\ 7^3 &\equiv 343 \equiv 1 \pmod{9}, \\\ 8^3 &\equiv 512 \equiv -1 \pmod{9}. \end{aligned} $$

Now, we need to determine $ 2002^{2002} \mod 9 $. First, we find $ 2002 \mod 9 $:

$$ 2002 \div 9 = 222 \text{ remainder } 4 \quad \Rightarrow \quad 2002 \equiv 4 \pmod{9}. $$

Thus,

$$ 2002^{2002} \equiv 4^{2002} \pmod{9}. $$

Next, we use Euler’s theorem, which states that $ a^{\phi(n)} \equiv 1 \pmod{n} $ for $ \gcd(a, n) = 1 $. Here, $ \phi(9) = 6 $, so:

$$ 4^6 \equiv 1 \pmod{9}. $$

We can reduce the exponent $ 2002 \mod 6 $:

$$ 2002 \div 6 = 333 \text{ remainder } 4 \quad \Rightarrow \quad 2002 \equiv 4 \pmod{6}. $$

Thus,

$$ 4^{2002} \equiv 4^4 \pmod{9}. $$

$$ 4^4 = 256. $$

Finally, we find $ 256 \mod 9 $:

$$ 256 \div 9 = 28 \text{ remainder } 4 \quad \Rightarrow \quad 256 \equiv 4 \pmod{9}. $$

Therefore,

$$ 2002^{2002} \equiv 4 \pmod{9}. $$

We need to find the smallest $ t $ such that the sum of $ t $ cubes is congruent to 4 modulo 9. The possible sums of cubes modulo 9 are:

$$ \begin{aligned} 0 &\equiv 0 \pmod{9}, \\\ 1 &\equiv 1 \pmod{9}, \\\ -1 &\equiv 8 \pmod{9}, \\\ 1 + 1 &\equiv 2 \pmod{9}, \\\ 1 + (-1) &\equiv 0 \pmod{9}, \\\ -1 + (-1) &\equiv -2 \equiv 7 \pmod{9}, \\\ 1 + 1 + 1 &\equiv 3 \pmod{9}, \\\ 1 + 1 + (-1) &\equiv 1 \pmod{9}, \\\ 1 + (-1) + (-1) &\equiv -1 \equiv 8 \pmod{9}, \\\ -1 + (-1) + (-1) &\equiv -3 \equiv 6 \pmod{9}, \\\ 1 + 1 + 1 + 1 &\equiv 4 \pmod{9}. \end{aligned} $$

We see that the sum of four cubes can be congruent to 4 modulo 9. Therefore, the smallest $ t $ is 4.

Thus, the smallest positive integer $ t $ such that there exist integers $ x_1, x_2, \ldots, x_t $ with

$$ x_1^3 + x_2^3 + \cdots + x_t^3 = 2002^{2002} $$

is

$$ \boxed{4}. $$

## Problem From IMO Shortlist 2022

**Problem**

For a positive integer $n$ we denote by $s(n)$ the sum of the digits of $n$. Let $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be a polynomial, where $n \geqslant 2$ and $a_i$ is a positive integer for all $ 0 \leqslant i \leqslant n-1$. Could it be the case that, for all positive integers $k$, $s(k)$ and $s(P(k))$ have the same parity?

**Qwen2-Math-72B-Instruct Solution**

To determine whether it is possible for $ s(k) $ and $ s(P(k)) $ to have the same parity for all positive integers $ k $, we need to analyze the behavior of the sum of the digits function $ s $ and the polynomial $ P(x) $.

First, recall that the sum of the digits of a number $ k $ modulo 2 is equivalent to $ k \mod 2 $. This is because the sum of the digits of $ k $ in base 10 is congruent to $ k \mod 9 $, and since $ 9 \equiv 1 \mod 2 $, the sum of the digits modulo 2 is the same as $ k \mod 2 $.

Given the polynomial $ P(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 $, we need to determine the parity of $ P(k) $ for any positive integer $ k $. Specifically, we need to check if $ P(k) \mod 2 $ is the same as $ k \mod 2 $.

Let’s consider the polynomial modulo 2:

$$ P(x) \equiv x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \mod 2. $$

Since $ a_i $ are positive integers, $ a_i \mod 2 $ is either 0 or 1. Let $ b_i = a_i \mod 2 $. Then we can rewrite the polynomial modulo 2 as:

$$ P(x) \equiv x^n + b_{n-1}x^{n-1} + \cdots + b_1x + b_0 \mod 2. $$

Now, we need to check the parity of $ P(k) $ for $ k \equiv 0 \mod 2 $ and $ k \equiv 1 \mod 2 $.

**Case 1: $ k \equiv 0 \mod 2 $**$$ P(0) \equiv b_0 \mod 2. $$

For $ s(P(0)) $ to have the same parity as $ s(0) $, $ b_0 $ must be 0. Therefore, $ a_0 $ must be even.

**Case 2: $ k \equiv 1 \mod 2 $**$$ P(1) \equiv 1^n + b_{n-1} \cdot 1^{n-1} + \cdots + b_1 \cdot 1 + b_0 \equiv 1 + b_{n-1} + \cdots + b_1 + b_0 \mod 2. $$

For $ s(P(1)) $ to have the same parity as $ s(1) $, $ 1 + b_{n-1} + \cdots + b_1 + b_0 $ must be odd. Since $ 1 $ is odd, the sum $ b_{n-1} + \cdots + b_1 + b_0 $ must be even.

From the above cases, we have:

- $ a_0 $ must be even.
- The sum $ b_{n-1} + \cdots + b_1 + b_0 $ must be even.

However, if $ a_0 $ is even, then $ b_0 = 0 $. This means the sum $ b_{n-1} + \cdots + b_1 $ must be even. But since $ b_i $ are either 0 or 1, the sum of $ b_{n-1} + \cdots + b_1 $ being even implies that there must be an even number of 1’s among $ b_{n-1}, \ldots, b_1 $.

Now, consider the polynomial $ P(x) = x^n + 1 $. For this polynomial:

- $ a_0 = 1 $ (odd), which contradicts our requirement that $ a_0 $ must be even.
- The sum $ b_{n-1} + \cdots + b_1 + b_0 = 1 $ (odd), which also contradicts our requirement that the sum must be even.

Therefore, it is impossible for $ s(k) $ and $ s(P(k)) $ to have the same parity for all positive integers $ k $.

Thus, the final answer is:

$$ \boxed{\text{No}} $$

## Problem From IMO 2022

**Problem**

Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying

$$ xf(y)+yf(x) \leq 2 $$

**Qwen2-Math-72B-Instruct Solution**

To find all functions $ f: \mathbb{R}^+ \to \mathbb{R}^+ $ such that for each $ x \in \mathbb{R}^+ $, there is exactly one $ y \in \mathbb{R}^+ $ satisfying

$$ xf(y) + yf(x) \leq 2, $$

we will proceed step by step.

**Step 1: Analyze the given inequality**
Given $ x \in \mathbb{R}^+ $, there exists a unique $ y \in \mathbb{R}^+ $ such that

$$ xf(y) + yf(x) \leq 2. $$

We need to explore the implications of this condition.

**Step 2: Consider the case $ x = y $**
Substitute $ y = x $ into the inequality:

$$ xf(x) + xf(x) = 2xf(x) \leq 2. $$

This simplifies to:

$$ xf(x) \leq 1. $$

Thus, for all $ x \in \mathbb{R}^+ $,

$$ f(x) \leq \frac{1}{x}. $$

**Step 3: Consider the uniqueness condition**
For each $ x \in \mathbb{R}^+ $, there is exactly one $ y \in \mathbb{R}^+ $ such that $ xf(y) + yf(x) \leq 2 $. We need to ensure that this $ y $ is unique.

**Step 4: Test the function $ f(x) = \frac{1}{x} $**
Consider the function $ f(x) = \frac{1}{x} $. Substitute $ f(x) = \frac{1}{x} $ and $ f(y) = \frac{1}{y} $ into the inequality:

$$ x \cdot \frac{1}{y} + y \cdot \frac{1}{x} = \frac{x}{y} + \frac{y}{x}. $$

We need:

$$ \frac{x}{y} + \frac{y}{x} \leq 2. $$

Using the AM-GM inequality, we know:

$$ \frac{x}{y} + \frac{y}{x} \geq 2. $$

Equality holds if and only if $ \frac{x}{y} = 1 $, which implies $ x = y $.

Thus, for $ f(x) = \frac{1}{x} $, the inequality $ xf(y) + yf(x) \leq 2 $ holds if and only if $ x = y $. This satisfies the condition that there is exactly one $ y $ for each $ x $.

**Step 5: Verify if other functions can satisfy the condition**
Suppose there exists another function $ g: \mathbb{R}^+ \to \mathbb{R}^+ $ such that $ g(x) \neq \frac{1}{x} $ for some $ x $. Then, for some $ x $, $ g(x) < \frac{1}{x} $ or $ g(x) > \frac{1}{x} $.

- If $ g(x) < \frac{1}{x} $, then $ xg(y) + yg(x) $ might not satisfy the inequality for all $ y $.
- If $ g(x) > \frac{1}{x} $, then $ xg(y) + yg(x) $ might exceed 2 for some $ y $.

Thus, any function other than $ f(x) = \frac{1}{x} $ would not satisfy the uniqueness condition.

**Conclusion**
The only function $ f: \mathbb{R}^+ \to \mathbb{R}^+ $ that satisfies the given condition is:

$$ \boxed{f(x) = \frac{1}{x}}. $$

## Problem from International Zhautykov Olympiad 2020

**Problem**

Let $Z$ be the set of all integers. Find all the function $f: Z \to Z$ such that

$$ f(4x+3y)=f(3x+y)+f(x+2y) $$

For all integers $x,y$

**Qwen2-Math-72B-Instruct Solution**

To find all functions $ f: \mathbb{Z} \to \mathbb{Z} $ that satisfy the functional equation

$$ f(4x + 3y) = f(3x + y) + f(x + 2y) $$

for all integers $ x $ and $ y $, we will proceed step by step.

**Step 1: Initial Substitution**

First, let’s substitute specific values for $ x $ and $ y $ to gain insight into the function.

Substitution 1: $ x = 0 $

$$ f(4 \cdot 0 + 3y) = f(3 \cdot 0 + y) + f(0 + 2y) $$

$$ f(3y) = f(y) + f(2y) $$

Substitution 2: $ y = 0 $

$$ f(4x + 3 \cdot 0) = f(3x + 0) + f(x + 2 \cdot 0) $$

$$ f(4x) = f(3x) + f(x) $$

**Step 2: Exploring Further Substitutions**

We now have two equations:

- $ f(3y) = f(y) + f(2y) $
- $ f(4x) = f(3x) + f(x) $

Substitution 3: $ y = 1 $ in the first equation

$$ f(3 \cdot 1) = f(1) + f(2 \cdot 1) $$

$$ f(3) = f(1) + f(2) $$

Substitution 4: $ x = 1 $ in the second equation

$$ f(4 \cdot 1) = f(3 \cdot 1) + f(1) $$

$$ f(4) = f(3) + f(1) $$

**Step 3: Using the Equations to Find a Pattern**

We have:

$$ f(3) = f(1) + f(2) $$

$$ f(4) = f(3) + f(1) $$

Substitute $ f(3) $ from the first equation into the second equation:

$$ f(4) = (f(1) + f(2)) + f(1) $$

$$ f(4) = 2f(1) + f(2) $$

**Step 4: Generalizing the Function**

We hypothesize that $ f $ might be a linear function. Let’s assume $ f(x) = kx $ for some integer $ k $.

*Verification*
Substitute $ f(x) = kx $ into the original functional equation:

$$ f(4x + 3y) = k(4x + 3y) $$

$$ f(3x + y) = k(3x + y) $$

$$ f(x + 2y) = k(x + 2y) $$

The original equation becomes:

$$ k(4x + 3y) = k(3x + y) + k(x + 2y) $$

$$ 4kx + 3ky = 3kx + ky + kx + 2ky $$

$$ 4kx + 3ky = 4kx + 3ky $$

This equation holds true for all $ x $ and $ y $, confirming that $ f(x) = kx $ is a solution.

**Conclusion**

The only functions $ f: \mathbb{Z} \to \mathbb{Z} $ that satisfy the given functional equation are of the form $ f(x) = kx $ for some integer $ k $.

Thus, the final answer is:

$$ \boxed{f(x) = kx \text{ for some integer } k} $$

## Problem From Baltic Way 2023

**Problem**

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$ f(f(x)+y)+xf(y)=f(xy+y)+f(x) $$

```
for reals $x, y$.
```

**Qwen2-Math-72B-Instruct Solution:**

To find all functions $ f: \mathbb{R} \rightarrow \mathbb{R} $ that satisfy the functional equation

$$ f(f(x) + y) + x f(y) = f(xy + y) + f(x), $$

we will proceed step by step.

**Step 1: Initial Substitution**

First, let’s substitute $ y = 0 $ into the equation:

$$ f(f(x) + 0) + x f(0) = f(x \cdot 0 + 0) + f(x). $$

This simplifies to:

$$ f(f(x)) + x f(0) = f(0) + f(x). $$

Rearranging terms, we get:

$$ f(f(x)) = f(x) + f(0) - x f(0). $$

**Step 2: Analyzing the Simplified Equation**

Let $ c = f(0) $. The equation becomes:

$$ f(f(x)) = f(x) + c - x c. $$

**Step 3: Substituting $ x = 0 $**

Next, substitute $ x = 0 $ into the original equation:

$$ f(f(0) + y) + 0 \cdot f(y) = f(0 \cdot y + y) + f(0). $$

This simplifies to:

$$ f(c + y) = f(y) + c. $$

**Step 4: Substituting $ y = -c $**

Substitute $ y = -c $ into the equation $ f(c + y) = f(y) + c $:

$$ f(c - c) = f(-c) + c. $$

This simplifies to:

$$ f(0) = f(-c) + c. $$

Since $ f(0) = c $, we have:

$$ c = f(-c) + c. $$

Subtracting $ c $ from both sides, we get:

$$ f(-c) = 0. $$

**Step 5: Substituting $ y = -f(x) $**

Substitute $ y = -f(x) $ into the original equation:

$$ f(f(x) - f(x)) + x f(-f(x)) = f(x(-f(x)) - f(x)) + f(x). $$

This simplifies to:

$$ f(0) + x f(-f(x)) = f(-x f(x) - f(x)) + f(x). $$

Since $ f(0) = c $, we have:

$$ c + x f(-f(x)) = f(-x f(x) - f(x)) + f(x). $$

**Step 6: Considering $ f(x) = 0 $**

Assume $ f(x) = 0 $ for all $ x $. Substitute $ f(x) = 0 $ into the original equation:

$$ f(0 + y) + x f(y) = f(xy + y) + f(x). $$

This simplifies to:

$$ f(y) + 0 = f(y) + 0, $$

which is always true.

**Step 7: Verifying $ f(x) = x $**

Assume $ f(x) = x $. Substitute $ f(x) = x $ into the original equation:

$$ f(x + y) + x f(y) = f(xy + y) + f(x). $$

This simplifies to:

$$ x + y + xy = xy + y + x, $$

which is always true.

**Conclusion**

The only functions that satisfy the given functional equation are $ f(x) = 0 $ and $ f(x) = x $. Therefore, the solutions are:

$$ \boxed{f(x) = 0 \text{ or } f(x) = x}. $$

## Problem From Lusophon Mathematical Olympiad 2023

**Problem**

A long time ago, there existed Martians with $ 3$ different colours: red, green and blue. As Mars was devastated by an intergalactic war, only $ 2$ Martians of each colours survived. In order to reconstruct the Martian population, they decided to use a machine that transforms two Martians of distinct colours into four Martians of colour different to the two initial ones. For example, if a red Martian and a blue Martian use the machine, they’ll be transformed into four green Martians.

a) Is it possible that, after using that machine finitely many times, we have $ 2022 $ red Martians, $ 2022$ green Martians and $ 2022$ blue Martians?

b) Is it possible that, after using that machine finitely many times, we have $ 2021$ red Martians, $ 2022$ green Martians and $ 2023$ blue Martians?

**Qwen2-Math-72B-Instruct Solution**

Let’s analyze the problem step by step.

**Part (a)**

We start with 2 red, 2 green, and 2 blue Martians. The machine transforms two Martians of distinct colors into four Martians of the third color. We need to determine if it is possible to end up with 2022 red, 2022 green, and 2022 blue Martians.

First, let’s consider the parity (odd or even nature) of the number of Martians of each color. Initially, we have:

- 2 red (even)
- 2 green (even)
- 2 blue (even)

Each transformation changes the parity of the number of Martians of the two initial colors and the parity of the number of Martians of the third color. Specifically:

- If we transform 1 red and 1 green Martian, we get 4 blue Martians. This changes the parity of red and green Martians from even to odd, and the parity of blue Martians from even to odd.
- If we transform 1 red and 1 blue Martian, we get 4 green Martians. This changes the parity of red and blue Martians from even to odd, and the parity of green Martians from even to odd.
- If we transform 1 green and 1 blue Martian, we get 4 red Martians. This changes the parity of green and blue Martians from even to odd, and the parity of red Martians from even to odd.

In all cases, the parity of the number of Martians of each color changes from even to odd or from odd to even. Therefore, after any number of transformations, the parity of the number of Martians of each color will always be the same as it was initially.

Since initially all three colors have an even number of Martians, after any number of transformations, the number of Martians of each color will remain even.

Given that 2022 is an even number, it is possible to have 2022 red, 2022 green, and 2022 blue Martians after a finite number of transformations.

Thus, the answer to part (a) is:

$$ \boxed{\text{Yes}} $$

**Part (b)**

We start again with 2 red, 2 green, and 2 blue Martians. We need to determine if it is possible to end up with 2021 red, 2022 green, and 2023 blue Martians.

As in part (a), we consider the parity of the number of Martians of each color. Initially, we have:

- 2 red (even)
- 2 green (even)
- 2 blue (even)

Each transformation changes the parity of the number of Martians of the two initial colors and the parity of the number of Martians of the third color. Therefore, after any number of transformations, the parity of the number of Martians of each color will always be the same as it was initially.

Since initially all three colors have an even number of Martians, after any number of transformations, the number of Martians of each color will remain even.

Given that 2021 is an odd number, it is impossible to have 2021 red Martians after any number of transformations. Similarly, 2023 is an odd number, making it impossible to have 2023 blue Martians.

Thus, the answer to part (b) is:

$$ \boxed{\text{No}} $$

## Problem From Balkan MO 2023

**Problem**

Joe and Penny play a game. Initially there are $ 5000$ stones in a pile, and the two players remove stones from the pile by making a sequence of moves. On the $ k$-th move, any number of stones between $ 1$ and $ k$ inclusive may be removed. Joe makes the odd-numbered moves and Penny makes the even-numbered moves. The player who removes the very last stone is the winner. Who wins if both players play perfectly?

**Qwen2-Math-72B-Instruct Solution**

To determine the winner of the game, we need to analyze the strategy and the sequence of moves. Let’s break down the problem step by step.

**Understanding the Game Rules:**- Initially, there are 5000 stones.
- Joe makes the first move (odd-numbered move), Penny makes the second move (even-numbered move), and so on.
- On the $ k $-th move, a player can remove between 1 and $ k $ stones.
- The player who removes the last stone wins.

**Key Insight:**- The game is symmetric in the sense that the number of stones a player can remove on their turn increases linearly with the move number.
- The total number of moves in the game is equal to the number of stones initially in the pile, which is 5000.

**Total Number of Moves:**- Since there are 5000 stones, there will be 5000 moves in total.
- Joe makes the odd-numbered moves: 1st, 3rd, 5th, …, 4999th.
- Penny makes the even-numbered moves: 2nd, 4th, 6th, …, 5000th.

**Determining the Winner:**- The last move (5000th move) is an even-numbered move.
- Therefore, Penny will make the last move.

**Conclusion:**- Since Penny makes the last move, she will be the one to remove the last stone.
- Hence, Penny wins the game if both players play perfectly.

Thus, the final answer is:

$$ \boxed{\text{Penny}} $$

## Problem From Math Odyssey

**Problem**

The school offers 10 elective courses, and each student can enroll in any number of courses. The director selects $k$ students, where although each student’s combination of courses is different, any two students have at least one course in common. At this point, it is found that any student outside these $k$ students cannot be classmates with these $k$ students regardless of how they enroll (having one course in common is enough to be classmates). Find the maximum number of $k$.

**Qwen2-Math-72B-Instruct Solution**

To solve this problem, we need to determine the maximum number of students $ k $ such that each student has a unique combination of elective courses, any two students share at least one course, and no student outside these $ k $ students can be classmates with all of them. This means that the set of students chosen must cover all possible courses in such a way that any new student would have to share a course with at least one of the chosen students.

First, let’s consider the total number of possible combinations of courses. Since there are 10 elective courses and each student can either take or not take each course, there are $ 2^{10} = 1024 $ possible combinations of courses. However, one of these combinations is the empty set (the student takes no courses), which is not a valid combination for our problem because it would not share any courses with any other student. Therefore, there are $ 1024 - 1 = 1023 $ non-empty combinations of courses.

We need to find the maximum number of students $ k $ such that any two students share at least one course, and no student outside these $ k $ students can be classmates with all of them. This is equivalent to finding the maximum number of vertices in a graph where each vertex represents a non-empty combination of courses, and there is an edge between two vertices if and only if the corresponding combinations share at least one course. The condition that no student outside these $ k $ students can be classmates with all of them means that the set of $ k $ students forms a dominating set in the graph.

The maximum size of such a set is the complement of the minimum size of an independent set in the graph. An independent set in this graph is a set of vertices such that no two vertices are adjacent, which means no two combinations share any course. The maximum size of an independent set in this graph is the number of combinations where each course is either taken or not taken by exactly one student, which is $ 2^9 = 512 $ (since we can choose to include or exclude each of the 9 remaining courses for each of the 10 courses).

Therefore, the minimum size of an independent set is 512, and the maximum size of a dominating set (which is the complement of the independent set) is $ 1023 - 512 + 1 = 511 $. However, we need to add 1 because the empty set is not included, so the correct calculation is $ 1023 - 511 = 512 $.

Thus, the maximum number of students $ k $ is $\boxed{512}$.

## Problem from USAMO 2010

**Problem**

Let $ABC$ be a triangle with $\angle A = 90^{\circ}$. Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$. Segments $BD$ and $CE$ meet at $I$. Determine whether or not it is possible for segments $AB$, $AC$, $BI$, $ID$, $CI$, $IE$ to all have integer lengths.

**Qwen2-Math-72B-Instruct Solution**

To determine whether it is possible for segments $AB$, $AC$, $BI$, $ID$, $CI$, and $IE$ to all have integer lengths in triangle $ABC$ with $\angle A = 90^\circ$, we start by analyzing the given conditions and properties of the triangle.

Given:

- $\angle A = 90^\circ$
- $\angle ABD = \angle DBC$
- $\angle ACE = \angle ECB$
- Segments $BD$ and $CE$ intersect at $I$

First, note that since $\angle A = 90^\circ$, triangle $ABC$ is a right triangle with $A$ as the right angle. Points $D$ and $E$ are such that $BD$ and $CE$ are angle bisectors of $\angle ABC$ and $\angle ACB$, respectively. Therefore, $I$ is the incenter of triangle $ABC$.

**Step 1: Coordinates and Lengths**
Assume $A$ is at the origin $(0, 0)$, $B$ is at $(a, 0)$, and $C$ is at $(0, b)$. The lengths of the sides are:

- $AB = a$
- $AC = b$
- $BC = \sqrt{a^2 + b^2}$

**Step 2: Incenter Coordinates**
The incenter $I$ of a right triangle with legs $a$ and $b$ and hypotenuse $\sqrt{a^2 + b^2}$ is located at:

$$ I = \left( \frac{a}{2}, \frac{b}{2} \right) $$

**Step 3: Lengths of Segments $BI$ and $CI$**
Using the distance formula, we calculate $BI$ and $CI$:

$$ BI = \sqrt{\left( a - \frac{a}{2} \right)^2 + \left( 0 - \frac{b}{2} \right)^2} = \sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2} = \frac{\sqrt{a^2 + b^2}}{2} $$

$$ CI = \sqrt{\left( 0 - \frac{a}{2} \right)^2 + \left( b - \frac{b}{2} \right)^2} = \sqrt{\left( \frac{a}{2} \right)^2 + \left( \frac{b}{2} \right)^2} = \frac{\sqrt{a^2 + b^2}}{2} $$

**Step 4: Lengths of Segments $ID$ and $IE$**
Since $D$ and $E$ are points on $AC$ and $AB$ respectively, and $I$ is the incenter, $ID$ and $IE$ are the distances from the incenter to the sides of the triangle. For a right triangle, these distances are given by:

$$ ID = IE = r = \frac{a + b - \sqrt{a^2 + b^2}}{2} $$

**Step 5: Integer Lengths**
For $AB$, $AC$, $BI$, $ID$, $CI$, and $IE$ to all be integers, the following must hold:

- $a$ and $b$ are integers
- $\frac{\sqrt{a^2 + b^2}}{2}$ is an integer
- $\frac{a + b - \sqrt{a^2 + b^2}}{2}$ is an integer

**Step 6: Analysis**
For $\frac{\sqrt{a^2 + b^2}}{2}$ to be an integer, $\sqrt{a^2 + b^2}$ must be an even integer. Let $\sqrt{a^2 + b^2} = 2k$ for some integer $k$. Then:

$$ a^2 + b^2 = 4k^2 $$

For $\frac{a + b - \sqrt{a^2 + b^2}}{2}$ to be an integer, $a + b - 2k$ must be even. Since $a$ and $b$ are integers, $a + b$ must be even.

**Conclusion**
Given the constraints, it is possible to find integer values for $a$ and $b$ such that all the conditions are satisfied. For example, if $a = 3$ and $b = 4$, then:

- $AB = 3$
- $AC = 4$
- $BC = 5$
- $BI = CI = \frac{5}{2} = 2.5$ (not an integer)

Thus, it is not possible for all segments $AB$, $AC$, $BI$, $ID$, $CI$, and $IE$ to have integer lengths simultaneously.

$$ \boxed{\text{No}} $$

## Problem from JBMO Shortlist 2011

**Problem**

Let $AD,BF$ and ${CE}$ be the altitudes of $\vartriangle ABC$. A line passing through ${D}$ and parallel to ${AB}$intersects the line ${EF}$at the point ${G}$. If ${H}$ is the orthocenter of $\vartriangle ABC$, find the angle ${\angle{CGH}}$.

**Qwen2-Math-72B-Instruct Solution**

To solve for the angle $\angle CGH$ in the given triangle $\triangle ABC$ with altitudes $AD$, $BF$, and $CE$ intersecting at the orthocenter $H$, and a line through $D$ parallel to $AB$ intersecting $EF$ at $G$, we will follow these steps:

**Identify Key Points and Properties:**- $D$ is the foot of the altitude from $A$ to $BC$.
- $E$ is the foot of the altitude from $B$ to $AC$.
- $F$ is the foot of the altitude from $C$ to $AB$.
- $H$ is the orthocenter of $\triangle ABC$.
- $G$ is the intersection of the line through $D$ parallel to $AB$ and the line $EF$.

**Use the Property of the Orthocenter:**- The orthocenter $H$ lies on the altitudes $AD$, $BF$, and $CE$.
- The points $D$, $E$, and $F$ are collinear with $H$ on the respective altitudes.

**Analyze the Parallel Line:**- Since $DG \parallel AB$, $\angle ADG = \angle DAB$ (corresponding angles).

**Consider the Cyclic Quadrilateral:**- The quadrilateral $AEHF$ is cyclic because $\angle AEH = \angle AFH = 90^\circ$.
- Therefore, $\angle EHF = 180^\circ - \angle EAF$.

**Use the Parallel Line Property Again:**- Since $DG \parallel AB$, $\angle DGE = \angle BAE$ (alterFnate interior angles).

**Relate Angles in Cyclic Quadrilateral:**- In $\triangle AEF$, $\angle EAF = \angle EHF$ because they are opposite angles in the cyclic quadrilateral $AEHF$.

**Determine $\angle CGH$:**- Since $G$ lies on $EF$ and $DG \parallel AB$, $\angle DGE = \angle BAE$.
- $\angle BAE = \angle EAF$ (as established).
- Therefore, $\angle DGE = \angle EAF$.

**Final Angle Calculation:**- Since $\angle EAF = \angle EHF$ and $\angle EHF = 90^\circ$ (as $H$ is the orthocenter and $E$ and $F$ are feet of perpendiculars), we have: $$ \angle CGH = \angle EHF = 90^\circ. $$

Thus, the angle $\angle CGH$ is $\boxed{90^\circ}$.

## 去除数据污染

我们在预训练和微调数据集上都进行了去污染处理。具体来说，对于预训练数据，我们针对数学数据集，包括 GSM8K、MATH，并移除与测试集有显著重叠的样本。我们移除了有13-gram重复且最长公共子序列比例大于0.6的训练样本。对于微调数据，我们移除了更多与 GSM8K、MATH、Aqua、SAT Math、OlympiadBench、College Math、AIME24、AMC23 等数据集有重叠的正例污染样本，使用了同样的过滤方法。

## 总结

这次我们发布的新模型系列 Qwen2-Math 专注于数学能力，构建于Qwen2的基础之上。我们的旗舰模型 Qwen2-Math-72B-Instruct 在数学相关任务上超越了诸如 GPT-4o 和 Claude 3.5 等专有模型。鉴于目前仅支持英文的限制，我们计划很快推出支持英文和中文的双语模型，并且多语言模型也在开发之中。此外，我们将持续增强模型解决复杂及具有挑战性的数学问题的能力。