AlgorithmicOps/aim-qwen2.5-7b-instruct-distil

---
size_categories: n<1K
dataset_info:
features:
- name: problem_id
dtype: string
- name: link
dtype: string
- name: problem
dtype: string
- name: solution
dtype: string
- name: letter
dtype: 'null'
- name: answer
dtype: float64
- name: year
dtype: int64
- name: season
dtype: 'null'
- name: competition
dtype: string
- name: version
dtype: string
- name: problem_number
dtype: int64
- name: category
dtype: string
- name: main_category
dtype: string
- name: subcategory
dtype: string
- name: generation
dtype: string
- name: distilabel_metadata
struct:
- name: raw_input_text_generation_0
list:
- name: content
dtype: string
- name: role
dtype: string
- name: raw_output_text_generation_0
dtype: string
- name: statistics_text_generation_0
struct:
- name: input_tokens
dtype: int64
- name: output_tokens
dtype: int64
- name: model_name
dtype: string
splits:
- name: train
num_bytes: 6985400
num_examples: 951
download_size: 3014024
dataset_size: 6985400
configs:
- config_name: default
data_files:
- split: train
path: data/train-*
tags:
- synthetic
- distilabel
- rlaif
---

<p align="left">
<a href="https://github.com/argilla-io/distilabel">
<img src="https://raw.githubusercontent.com/argilla-io/distilabel/main/docs/assets/distilabel-badge-light.png" alt="Built with Distilabel" width="200" height="32"/>
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# Dataset Card for aim-qwen2.5-7b-instruct-distil

This dataset has been created with [distilabel](https://distilabel.argilla.io/).

## Dataset Summary

This dataset contains a `pipeline.yaml` which can be used to reproduce the pipeline that generated it in distilabel using the `distilabel` CLI:

```console
distilabel pipeline run --config "https://huggingface.co/datasets/AlgorithmicOps/aim-qwen2.5-7b-instruct-distil/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/AlgorithmicOps/aim-qwen2.5-7b-instruct-distil/raw/main/pipeline.yaml"
```

## Dataset structure

The examples have the following structure per configuration:

<details><summary> Configuration: default </summary><hr>

```json
{
"answer": 809.0,
"category": "Discrete Mathematics: Combinatorial Game Theory",
"competition": "AIME",
"distilabel_metadata": {
"raw_input_text_generation_0": [
{
"content": "You will be given a problem. Please reason step by step, and put your final answer within \boxed{}:\nAlice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for which there exists a strategy for Bob that guarantees that Bob will win the game regardless of Alice\u0027s play.",
"role": "user"
}
],
"raw_output_text_generation_0": "To solve this problem, we need to analyze the game from the perspective of winning and losing positions. A winning position is one from which the player can force a win, and a losing position is one from which any move leaves the opponent in a winning position.\n\n### Step 1: Identify Basic Winning and Losing Positions\n- If there are 0 tokens left, the player whose turn it is loses (since they cannot make a move).\n- If there is 1 token, the player can take it and win.\n- If there are 4 tokens, the player can take all 4 and win.\n\n### Step 2: Identify Patterns\nWe need to determine the pattern of winning and losing positions for larger numbers of tokens.\n\n- If there are 2 or 3 tokens, the player can leave the opponent with 1 token, forcing a win.\n- If there are 5 tokens, the player can take 1 token, leaving 4 tokens (a winning position for the opponent).\n\n### Step 3: Generalize the Pattern\nWe can see that positions that are multiples of 5 (i.e., 5, 10, 15, ...) are losing positions because any move from these positions leaves the opponent in a winning position (1, 2, 3, or 4 tokens).\n\n### Step 4: Count Losing Positions\nWe need to count how many numbers from 1 to 2024 are multiples of 5.\n\n- The smallest multiple of 5 is 5.\n- The largest multiple of 5 less than or equal to 2024 is 2020 (since \\(2020 = 5 \\times 404\\)).\n\n### Step 5: Calculate the Number of Multiples of 5\nThe sequence of multiples of 5 from 5 to 2020 is an arithmetic sequence where the first term \\(a = 5\\) and the common difference \\(d = 5\\).\n\nThe \\(n\\)-th term of an arithmetic sequence is given by:\n\\[ a_n = a + (n-1)d \\]\n\nSetting \\(a_n = 2020\\):\n\\[ 2020 = 5 + (n-1) \\cdot 5 \\]\n\\[ 2020 = 5 + 5n - 5 \\]\n\\[ 2020 = 5n \\]\n\\[ n = 404 \\]\n\n### Step 6: Conclusion\nThere are 404 multiples of 5 between 1 and 2024. These are the losing positions for Bob. Therefore, there are 404 positive integers \\(n\\) less than or equal to 2024 for which Bob has a winning strategy if he plays optimally.\n\n\\[\n\\boxed{404}\n\\]",
"statistics_text_generation_0": {
"input_tokens": 152,
"output_tokens": 591
}
},
"generation": "To solve this problem, we need to analyze the game from the perspective of winning and losing positions. A winning position is one from which the player can force a win, and a losing position is one from which any move leaves the opponent in a winning position.\n\n### Step 1: Identify Basic Winning and Losing Positions\n- If there are 0 tokens left, the player whose turn it is loses (since they cannot make a move).\n- If there is 1 token, the player can take it and win.\n- If there are 4 tokens, the player can take all 4 and win.\n\n### Step 2: Identify Patterns\nWe need to determine the pattern of winning and losing positions for larger numbers of tokens.\n\n- If there are 2 or 3 tokens, the player can leave the opponent with 1 token, forcing a win.\n- If there are 5 tokens, the player can take 1 token, leaving 4 tokens (a winning position for the opponent).\n\n### Step 3: Generalize the Pattern\nWe can see that positions that are multiples of 5 (i.e., 5, 10, 15, ...) are losing positions because any move from these positions leaves the opponent in a winning position (1, 2, 3, or 4 tokens).\n\n### Step 4: Count Losing Positions\nWe need to count how many numbers from 1 to 2024 are multiples of 5.\n\n- The smallest multiple of 5 is 5.\n- The largest multiple of 5 less than or equal to 2024 is 2020 (since \\(2020 = 5 \\times 404\\)).\n\n### Step 5: Calculate the Number of Multiples of 5\nThe sequence of multiples of 5 from 5 to 2020 is an arithmetic sequence where the first term \\(a = 5\\) and the common difference \\(d = 5\\).\n\nThe \\(n\\)-th term of an arithmetic sequence is given by:\n\\[ a_n = a + (n-1)d \\]\n\nSetting \\(a_n = 2020\\):\n\\[ 2020 = 5 + (n-1) \\cdot 5 \\]\n\\[ 2020 = 5 + 5n - 5 \\]\n\\[ 2020 = 5n \\]\n\\[ n = 404 \\]\n\n### Step 6: Conclusion\nThere are 404 multiples of 5 between 1 and 2024. These are the losing positions for Bob. Therefore, there are 404 positive integers \\(n\\) less than or equal to 2024 for which Bob has a winning strategy if he plays optimally.\n\n\\[\n\\boxed{404}\n\\]",
"letter": null,
"link": "https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_3",
"main_category": "Discrete Mathematics",
"model_name": "Qwen/Qwen2.5-7B-Instruct",
"problem": "Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for which there exists a strategy for Bob that guarantees that Bob will win the game regardless of Alice\u0027s play.",
"problem_id": "f098258808bd480c688d1a313d743992",
"problem_number": 3,
"season": null,
"solution": "Let\u0027s first try some experimentation. Alice obviously wins if there is one coin. She will just take it and win. If there are 2 remaining, then Alice will take one and then Bob will take one, so Bob wins. If there are $3$ , Alice will take $1$ , Bob will take one, and Alice will take the final one. If there are $4$ , Alice will just remove all $4$ at once. If there are $5$ , no matter what Alice does, Bob can take the final coins in one try. Notice that Alice wins if there are $1$ $3$ , or $4$ coins left. Bob wins if there are $2$ or $5$ coins left.\nAfter some thought, you may realize that there is a strategy for Bob. If there is n is a multiple of $5$ , then Bob will win. The reason for this is the following: Let\u0027s say there are a multiple of $5$ coins remaining in the stack. If Alice takes $1$ , Bob will take $4$ , and there will still be a multiple of $5$ . If Alice takes $4$ , Bob will take $1$ , and there will still be a multiple of $5$ . This process will continue until you get $0$ coins left. For example, let\u0027s say there are $205$ coins. No matter what Alice does, Bob can simply just do the complement. After each of them make a turn, there will always be a multiple of $5$ left. This will continue until there are $5$ coins left, and Bob will end up winning.\nAfter some more experimentation, you\u0027ll realize that any number that is congruent to $2$ mod $5$ will also work. This is because Bob can do the same strategy, and when there are $2$ coins left, Alice is forced to take $1$ and Bob takes the final coin. For example, let\u0027s say there are $72$ coins. If Alice takes $1$ , Bob will take $4$ . If Alice takes $4$ , Bob will take $1$ . So after they each make a turn, the number will always be equal to $2$ mod $5$ . Eventually, there will be only $2$ coins remaining, and we\u0027ve established that Alice will simply take $1$ and Bob will take the final coin.\nSo we have to find the number of numbers less than or equal to $2024$ that are either congruent to $0$ mod $5$ or $2$ mod $5$ . There are $404$ numbers in the first category: $5, 10, 15, \\dots, 2020$ . For the second category, there are $405$ numbers. $2, 7, 12, 17, \\dots, 2022$ . So the answer is $404 + 405 = \\boxed{809}$",
"subcategory": " Combinatorial Game Theory",
"version": "I",
"year": 2024
}
```

This subset can be loaded as:

```python
from datasets import load_dataset

ds = load_dataset("AlgorithmicOps/aim-qwen2.5-7b-instruct-distil", "default")
```

Or simply as it follows, since there's only one configuration and is named `default`:

```python
from datasets import load_dataset

ds = load_dataset("AlgorithmicOps/aim-qwen2.5-7b-instruct-distil")
```

</details>