AlgorithmicOps/r1-7b-distil

---
size_categories: n<1K
dataset_info:
features:
- name: problem_id
dtype: string
- name: link
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- name: problem
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- name: solution
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- name: letter
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- name: answer
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- name: year
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- name: season
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- name: competition
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- name: version
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- name: problem_number
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- name: category
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- name: main_category
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- name: subcategory
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- name: generation
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- 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
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- name: output_tokens
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- name: model_name
dtype: string
splits:
- name: train
num_bytes: 48515507
num_examples: 951
download_size: 20101605
dataset_size: 48515507
configs:
- config_name: default
data_files:
- split: train
path: data/train-*
tags:
- synthetic
- distilabel
- rlaif
---

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# Dataset Card for r1-7b-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/r1-7b-distil/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/AlgorithmicOps/r1-7b-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": "Alright, so I came across this interesting game theory problem where Alice and Bob take turns removing either 1 or 4 tokens from a stack of n tokens. The person who removes the last token wins. The question is asking for the number of positive integers n \u2264 2024 where Bob can force a win, no matter how Alice plays. Hmm, okay, let me try to figure this out step by step.\n\nFirst, I remember in game theory problems like this, it\u0027s often useful to analyze smaller cases and look for patterns or periodicity. So, let me start by considering small values of n and determine whether Alice or Bob can win.\n\nLet\u2019s list out n from 1 upwards and see the outcome for each:\n\n- **n = 1**: Alice can take the 1 token and win immediately. So, Bob can\u0027t win here.\n- **n = 2**: Alice can take 1 token, leaving 1 for Bob, who then takes it and wins. Wait, no, if Alice takes 1, Bob is left with 1 and takes it. But actually, if n=2, Alice can take 1, leaving 1, so Bob takes the last one and wins. Wait, so does Bob win here? Or is it that Alice can force a win? Hmm, let me think.\n\nWait, no. If Alice takes 1 token from n=2, she leaves 1 token, which Bob can take and win. So, does that mean Bob can win when n=2? But wait, Alice could also take 2 tokens, but in the problem, she can only take 1 or 4. Since n=2, she can only take 1. So, Alice must take 1, leaving 1 for Bob, who then takes it and wins. So, Bob can win when n=2.\n\nWait, hold on, but the rules say each player can remove either 1 or 4 tokens. So, for n=2, Alice can only remove 1 token because she can\u0027t remove 4. So, she must leave 1 token for Bob, who then wins. So, n=2 is a winning position for Bob.\n\nWait, that seems a bit counterintuitive because usually, the player who can force the opponent into a losing position wins. So, if n=2, Alice is forced to leave Bob with 1, meaning Bob can win, so n=2 is a losing position for Alice and a winning position for Bob.\n\nWait, no, that can\u0027t be. If n=2 is a losing position for Alice, then Bob can win. But actually, in this case, the person who takes the last token wins, so if Alice takes 1, Bob takes the last one and wins. So, yes, n=2 is a Bob win.\n\nWait, maybe I should think in terms of winning (W) and losing (L) positions. A position is a winning position if the current player can force a win, and a losing position if no matter what the current player does, the opponent can win.\n\nSo, let\u0027s define:\n\n- n=0: This is a losing position because there are no tokens to take. But in our problem, n starts from 1, so maybe we can ignore n=0.\n\nBut for n=1: Alice can take 1 and win, so n=1 is a W position.\n\nn=2: Alice can take 1, leaving n=1 for Bob, who can then take it and win. So, n=2 is an L position for Alice, meaning Bob can win. Wait, n=2 is an L position for the current player (Alice), so it\u0027s a W position for Bob.\n\nWait, maybe I should define the positions based on whose turn it is. But perhaps it\u0027s simpler to think in terms of Grundy numbers or something, but maybe not necessary here.\n\nAlternatively, let\u0027s think recursively.\n\nFor each n, the current player can take 1 or 4 tokens. If either of those moves leads to the opponent being in a losing position, then the current position is a winning position. If all moves lead to the opponent being in a winning position, then the current position is losing.\n\nSo, let\u0027s try to define W(n) as true if the current player can force a win from n tokens, and false otherwise.\n\nBase cases:\n- If n=0: current player loses (since they can\u0027t move), so L.\n- If n=1: current player takes 1 and wins, so W.\n- If n=2: current player can only take 1, leaving n=1 for the opponent, who can then win. So, current player loses, so L.\n- If n=3: current player can take 1, leaving n=2 (which is L for the opponent). So, current player can win by taking 1, making n=2 for the opponent, who then loses. So, W.\n- If n=4: current player can take 1 or 4.\n - Taking 1 leaves n=3, which is W for the opponent.\n - Taking 4 leaves n=0, which is L for the opponent.\n So, current player can win by taking 4, so W.\n- If n=5: current player can take 1, leaving n=4 (which is W for opponent), or take 4, leaving n=1 (which is W for opponent). So, both moves leave opponent in W positions, so n=5 is L.\n- If n=6: current player can take 1, leaving n=5 (L for opponent), so current player can win by taking 1. So, W.\n- If n=7: current player can take 1, leaving n=6 (W for opponent), or take 4, leaving n=3 (W for opponent). So, both moves leave opponent in W positions, so n=7 is L.\n- If n=8: current player can take 1, leaving n=7 (L for opponent), so current player can win by taking 1. So, W.\n- If n=9: current player can take 1, leaving n=8 (W for opponent), or take 4, leaving n=5 (L for opponent). So, current player can win by taking 4. So, W.\n- If n=10: current player can take 1, leaving n=9 (W for opponent), or take 4, leaving n=6 (W for opponent). So, both moves leave opponent in W positions, so n=10 is L.\n \nHmm, so let\u0027s list out the W and L positions from n=1 to n=10:\n\nn: 1 2 3 4 5 6 7 8 9 10\n\nW/L: W L W W W L W L W L\n\nWait, let me check that again:\n\nn=1: W\n\nn=2: current player takes 1, opponent gets n=1 (W for opponent), so current player loses. So, n=2: L\n\nn=3: current player takes 1, opponent gets n=2 (L for opponent), so current player can win. So, n=3: W\n\nn=4: current player can take 1 or 4. Taking 1 leaves n=3 (W for opponent), taking 4 leaves n=0 (L for opponent). So, current player can win by taking 4. So, n=4: W\n\nn=5: current player takes 1, opponent gets n=4 (W); takes 4, opponent gets n=1 (W). So, both moves leave opponent in W positions, so n=5: L\n\nn=6: current player takes 1, opponent gets n=5 (L). So, current player can win. So, n=6: W\n\nn=7: current player takes 1, opponent gets n=6 (W); takes 4, opponent gets n=3 (W). So, both moves leave opponent in W positions, so n=7: L\n\nn=8: current player takes 1, opponent gets n=7 (L). So, current player can win. So, n=8: W\n\nn=9: current player takes 1, opponent gets n=8 (W); takes 4, opponent gets n=5 (L). So, current player can win by taking 4. So, n=9: W\n\nn=10: current player takes 1, opponent gets n=9 (W); takes 4, opponent gets n=6 (W). So, both moves leave opponent in W positions, so n=10: L\n\nSo, the pattern seems to be L at n=2,5,7,10,... Let me write down the L positions so far: 2,5,7,10,...\n\nWait, let me check n=11:\n\nn=11: current player can take 1, leaving n=10 (L for opponent), so current player can win by taking 1. So, n=11: W\n\nn=12: current player can take 1, leaving n=11 (W); take 4, leaving n=8 (W). So, both moves leave opponent in W positions. So, n=12: L\n\nn=13: current player can take 1, leaving n=12 (L). So, current player can win by taking 1. So, n=13: W\n\nn=14: current player can take 1, leaving n=13 (W); take 4, leaving n=10 (L). So, current player can win by taking 4. So, n=14: W\n\nn=15: current player can take 1, leaving n=14 (W); take 4, leaving n=11 (W). So, both moves leave opponent in W positions. So, n=15: L\n\nn=16: current player can take 1, leaving n=15 (L). So, current player can win by taking 1. So, n=16: W\n\nn=17: current player can take 1, leaving n=16 (W); take 4, leaving n=13 (W). So, both moves leave opponent in W positions. So, n=17: L\n\nn=18: current player can take 1, leaving n=17 (L). So, current player can win by taking 1. So, n=18: W\n\nn=19: current player can take 1, leaving n=18 (W); take 4, leaving n=15 (L). So, current player can win by taking 4. So, n=19: W\n\nn=20: current player can take 1, leaving n=19 (W); take 4, leaving n=16 (W). Both moves leave opponent in W positions. So, n=20: L\n\nSo, the L positions so far are: 2,5,7,10,12,15,17,20,...\n\nLet me see if I can find a pattern here. Let\u0027s list them:\n\n2,5,7,10,12,15,17,20,22,25,27,30,...\n\nWait, let\u0027s compute the differences between consecutive L positions:\n\n5-2=3\n\n7-5=2\n\n10-7=3\n\n12-10=2\n\n15-12=3\n\n17-15=2\n\n20-17=3\n\n22-20=2\n\n25-22=3\n\n27-25=2\n\n30-27=3\n\nSo, the pattern is alternating differences of 3 and 2. So, starting at 2, then adding 3, then 2, then 3, etc.\n\nSo, the L positions are numbers congruent to 2 mod 5 and 0 mod 5? Wait, let me check:\n\n2 mod 5 is 2\n\n5 mod 5 is 0\n\n7 mod 5 is 2\n\n10 mod 5 is 0\n\n12 mod 5 is 2\n\n15 mod 5 is 0\n\n17 mod 5 is 2\n\n20 mod 5 is 0\n\n22 mod 5 is 2\n\n25 mod 5 is 0\n\n27 mod 5 is 2\n\n30 mod 5 is 0\n\nYes, so it alternates between numbers that are 0 mod 5 and 2 mod 5. So, the losing positions are numbers that are 0 or 2 mod 5.\n\nWait, hold on. Let me check:\n\nn=2: 2 mod 5=2, yes.\n\nn=5: 5 mod5=0.\n\nn=7: 7 mod5=2.\n\nn=10:10 mod5=0.\n\nn=12:12 mod5=2.\n\nn=15:15 mod5=0.\n\nn=17:17 mod5=2.\n\nn=20:20 mod5=0.\n\nYes, so the losing positions are numbers that are 0 or 2 mod5.\n\nSo, for n\u22610 mod5 or n\u22612 mod5, it\u0027s a losing position.\n\nWait, but when n=5, which is 0 mod5, and n=2, which is 2 mod5, both are losing positions.\n\nBut let\u0027s test n=25 and n=27, which should be losing positions.\n\nn=25: If it\u0027s 0 mod5, then it\u0027s L.\n\nn=26: current player can take 1, leaving n=25 (L). So, n=26: W\n\nn=27: current player can take 1, leaving n=26 (W); take 4, leaving n=23. Wait, let me check n=23.\n\nWait, maybe I should check n=23.\n\nWait, maybe I need to test further.\n\nWait, perhaps my initial analysis was correct up to n=20, but n=23 may not follow the same pattern. Let me compute n=21 to n=25.\n\nWait, n=21:\n\nn=21: current player can take 1, leaving n=20 (L). So, n=21: W\n\nn=22: current player can take 1, leaving n=21 (W); take 4, leaving n=18 (W). So, both moves leave opponent in W positions, so n=22: L\n\nn=23: current player can take 1, leaving n=22 (L). So, n=23: W\n\nn=24: current player can take 1, leaving n=23 (W); take 4, leaving n=20 (L). So, current player can win by taking 4, leaving n=20 (L). So, n=24: W\n\nn=25: current player can take 1, leaving n=24 (W); take 4, leaving n=21 (W). So, both moves leave opponent in W positions, so n=25: L\n\nSo, n=25: L, which is 0 mod5.\n\nSimilarly, n=26: W, as before.\n\nn=27: current player can take 1, leaving n=26 (W); take 4, leaving n=23 (W). So, both moves leave opponent in W positions, so n=27: L\n\nYes, so n=27 is L, which is 2 mod5.\n\nSo, the pattern seems consistent.\n\nTherefore, the losing positions are n \u22610 mod5 and n\u22612 mod5.\n\nTherefore, for Bob to have a winning strategy, n must be a losing position for Alice, i.e., n \u22610 or 2 mod5.\n\nWait, but hold on. The way the problem is stated, Alice goes first. So, if n is a losing position for Alice, meaning that no matter what she does, Bob can win. So, yes, Bob can win when n is a losing position for Alice, which are n\u22610 or 2 mod5.\n\nTherefore, the number of positive integers n \u22642024 where Bob can force a win is equal to the number of n \u22642024 such that n \u22610 or 2 mod5.\n\nSo, now, we need to compute the count of numbers \u22642024 that are \u22610 or 2 mod5.\n\nLet me compute how many numbers \u22642024 are \u22610 mod5 and how many are \u22612 mod5.\n\nFirst, for numbers \u22610 mod5:\n\nThe smallest is 5, then 10,15,... up to the largest \u22642024.\n\nWhat\u0027s the largest multiple of 5 \u22642024? 2024 \u00f75=404.8, so 404\u00d75=2020.\n\nSo, numbers are 5,10,...,2020.\n\nNumber of terms: 2020 /5=404.\n\nSimilarly, numbers \u22612 mod5:\n\nThese are 2,7,12,..., up to the largest \u22642024.\n\nLet me find the largest number \u22612 mod5 \u22642024.\n\n2024 \u00f75=404.8, so 404\u00d75=2020, then 2020+2=2022.\n\nSo, the sequence is 2,7,12,...,2022.\n\nNumber of terms: Let me compute how many terms.\n\nFirst term:2, last term:2022, common difference:5.\n\nNumber of terms= ((2022 -2)/5)+1= (2020/5)+1=404+1=405.\n\nTherefore, total numbers \u22610 mod5:404, \u22612 mod5:405.\n\nTotal numbers:404 +405=809.\n\nBut wait, n must be positive integers, so n starts at 1. However, n=0 is not considered as per the problem.\n\nWait, in our analysis, n=0 is losing, but n starts at 1. So, n=0 isn\u0027t counted. So, the numbers \u22610 mod5 are from 5 to2020, which is 404 numbers, and numbers \u22612 mod5 are from2 to2022, which is405 numbers.\n\nTherefore, total numbers where Bob can win:404 +405=809.\n\nBut let me verify: 5\u00d7404=2020, correct. For 2 mod5, 2022=5\u00d7404 +2, so 405 numbers. So, yes, total is809.\n\nWait, but hold on. Let me think again.\n\nIf we count n from1 to2024, inclusive.\n\nThe number of integers \u22610 mod5 is floor(2024/5)=404 (since 5\u00d7404=2020).\n\nThe number of integers \u22612 mod5 is floor((2024 -2)/5) +1= floor(2022/5)+1=404 +1=405, as above.\n\nSo, total is404 +405=809.\n\nBut wait, let me check for n=1:1\u22611 mod5, which is a winning position, so Alice can win.\n\nn=2: losing position, Bob can win.\n\nn=3: winning, Alice can win.\n\nn=4: winning, Alice can win.\n\nn=5: losing, Bob can win.\n\nSo, the first few numbers where Bob can win are2,5,7,10,12,15,17,20,22,25,27,...\n\nWhich is exactly numbers \u22610 or2 mod5.\n\nSo, the count is809.\n\nBut hold on, wait. Let\u0027s make sure that for n=2022, which is a losing position, it\u0027s included since2022\u22642024.\n\nYes, 2022 is included.\n\nSimilarly, n=2025 would be the next multiple of5, but since2025\u003e2024, it\u0027s not included.\n\nSimilarly, n=2027 would be the next number \u22612 mod5, but2027\u003e2024, so not included.\n\nTherefore, total numbers where Bob can win is809.\n\nWait, but let me just compute 404 +405:\n\n404 +405=809.\n\nBut wait, 404 +405=809, correct.\n\nBut wait, the total number of integers from1 to2024 is2024.\n\nNumber of losing positions is809, meaning the remaining are winning positions.\n\nBut 2024 -809=1215.\n\nBut wait, 2024 divided by5 is404.8, so 404 numbers for each residue, but since2024 is not a multiple of5, some residues have one more number.\n\nWait, actually, in the range1 to2024, residues mod5:\n\nEach residue from0 to4 has either404 or405 numbers.\n\nSpecifically, residues0 and2 have404 and405?\n\nWait, no, let me compute.\n\nTotal numbers:2024.\n\nNumber per residue: floor(2024/5)=404 with remainder4.\n\nSo, residues1,2,3,4 will have405 numbers each, and residue0 has404.\n\nWait, no, actually, when you distribute2024 into5 residue classes, each class will havefloor(2024/5)=404, and the remainder is2024%5=2024 -5\u00d7404=2024-2020=4. So, residues1,2,3,4 will have1 more each, so405 each, and residue0 has404.\n\nTherefore, the number of numbers \u22610 mod5 is404, and numbers\u22611,2,3,4 mod5 are405 each.\n\nWait, but in our case, the losing positions are\u22610 and2 mod5.\n\nSo, number of losing positions=404 +405=809.\n\nHence, the count is809.\n\nTherefore, the number of positive integers n \u22642024 for which Bob can force a win is809.\n\nBut before finalizing, let me cross-verify with another approach.\n\nAnother way to think is that the losing positions are periodic with period5, repeating every5 numbers, with two losing positions in each period: positions2 and5, which are2 and0 mod5.\n\nSo, in each block of5 numbers, there are2 losing positions.\n\nTotal number of blocks in2024 numbers:2024//5=404 full blocks, each contributing2 losing positions:404\u00d72=808.\n\nThen, the remaining numbers after404\u00d75=2020 numbers are2021 to2024, which are4 numbers.\n\nIn these4 numbers, check if any are losing positions.\n\nThe losing positions are\u22610 or2 mod5.\n\n2021\u22611 mod5, 2022\u22612 mod5,2023\u22613 mod5,2024\u22614 mod5.\n\nSo, among2021-2024, only2022 is a losing position.\n\nTherefore, total losing positions=404\u00d72 +1=809.\n\nYes, same result.\n\nTherefore, I can be confident that the answer is809.\n\n**Final Answer**\nThe number of positive integers \\( n \\) less than or equal to 2024 for which Bob can force a win is \\(\\boxed{809}\\).\n\u003c/think\u003e\n\nTo determine the number of positive integers \\( n \\leq 2024 \\) for which Bob can force a win, we analyze the game where Alice and Bob take turns removing either 1 or 4 tokens from a stack. The player who removes the last token wins.\n\nWe define positions as winning (W) or losing (L) based on whether the current player can force a win. Through analysis, we find that the losing positions (where the current player cannot force a win) are those where \\( n \\equiv 0 \\pmod{5} \\) or \\( n \\equiv 2 \\pmod{5} \\).\n\nTo find the number of such losing positions \\( n \\leq 2024 \\):\n\n1. Calculate the number of integers \\( n \\leq 2024 \\) such that \\( n \\equiv 0 \\pmod{5} \\):\n - The largest multiple of 5 less than or equal to 2024 is 2020.\n - The number of such integers is \\( \\frac{2020}{5} = 404 \\).\n\n2. Calculate the number of integers \\( n \\leq 2024 \\) such that \\( n \\equiv 2 \\pmod{5} \\):\n - The largest such integer less than or equal to 2024 is 2022.\n - The number of such integers is \\( \\frac{2022 - 2}{5} + 1 = 405 \\).\n\nAdding these, the total number of losing positions is \\( 404 + 405 = 809 \\).\n\nThus, the number of positive integers \\( n \\leq 2024 \\) for which Bob can force a win is \\(\\boxed{809}\\).",
"statistics_text_generation_0": {
"input_tokens": 129,
"output_tokens": 5678
}
},
"generation": "Alright, so I came across this interesting game theory problem where Alice and Bob take turns removing either 1 or 4 tokens from a stack of n tokens. The person who removes the last token wins. The question is asking for the number of positive integers n \u2264 2024 where Bob can force a win, no matter how Alice plays. Hmm, okay, let me try to figure this out step by step.\n\nFirst, I remember in game theory problems like this, it\u0027s often useful to analyze smaller cases and look for patterns or periodicity. So, let me start by considering small values of n and determine whether Alice or Bob can win.\n\nLet\u2019s list out n from 1 upwards and see the outcome for each:\n\n- **n = 1**: Alice can take the 1 token and win immediately. So, Bob can\u0027t win here.\n- **n = 2**: Alice can take 1 token, leaving 1 for Bob, who then takes it and wins. Wait, no, if Alice takes 1, Bob is left with 1 and takes it. But actually, if n=2, Alice can take 1, leaving 1, so Bob takes the last one and wins. Wait, so does Bob win here? Or is it that Alice can force a win? Hmm, let me think.\n\nWait, no. If Alice takes 1 token from n=2, she leaves 1 token, which Bob can take and win. So, does that mean Bob can win when n=2? But wait, Alice could also take 2 tokens, but in the problem, she can only take 1 or 4. Since n=2, she can only take 1. So, Alice must take 1, leaving 1 for Bob, who then takes it and wins. So, Bob can win when n=2.\n\nWait, hold on, but the rules say each player can remove either 1 or 4 tokens. So, for n=2, Alice can only remove 1 token because she can\u0027t remove 4. So, she must leave 1 token for Bob, who then wins. So, n=2 is a winning position for Bob.\n\nWait, that seems a bit counterintuitive because usually, the player who can force the opponent into a losing position wins. So, if n=2, Alice is forced to leave Bob with 1, meaning Bob can win, so n=2 is a losing position for Alice and a winning position for Bob.\n\nWait, no, that can\u0027t be. If n=2 is a losing position for Alice, then Bob can win. But actually, in this case, the person who takes the last token wins, so if Alice takes 1, Bob takes the last one and wins. So, yes, n=2 is a Bob win.\n\nWait, maybe I should think in terms of winning (W) and losing (L) positions. A position is a winning position if the current player can force a win, and a losing position if no matter what the current player does, the opponent can win.\n\nSo, let\u0027s define:\n\n- n=0: This is a losing position because there are no tokens to take. But in our problem, n starts from 1, so maybe we can ignore n=0.\n\nBut for n=1: Alice can take 1 and win, so n=1 is a W position.\n\nn=2: Alice can take 1, leaving n=1 for Bob, who can then take it and win. So, n=2 is an L position for Alice, meaning Bob can win. Wait, n=2 is an L position for the current player (Alice), so it\u0027s a W position for Bob.\n\nWait, maybe I should define the positions based on whose turn it is. But perhaps it\u0027s simpler to think in terms of Grundy numbers or something, but maybe not necessary here.\n\nAlternatively, let\u0027s think recursively.\n\nFor each n, the current player can take 1 or 4 tokens. If either of those moves leads to the opponent being in a losing position, then the current position is a winning position. If all moves lead to the opponent being in a winning position, then the current position is losing.\n\nSo, let\u0027s try to define W(n) as true if the current player can force a win from n tokens, and false otherwise.\n\nBase cases:\n- If n=0: current player loses (since they can\u0027t move), so L.\n- If n=1: current player takes 1 and wins, so W.\n- If n=2: current player can only take 1, leaving n=1 for the opponent, who can then win. So, current player loses, so L.\n- If n=3: current player can take 1, leaving n=2 (which is L for the opponent). So, current player can win by taking 1, making n=2 for the opponent, who then loses. So, W.\n- If n=4: current player can take 1 or 4.\n - Taking 1 leaves n=3, which is W for the opponent.\n - Taking 4 leaves n=0, which is L for the opponent.\n So, current player can win by taking 4, so W.\n- If n=5: current player can take 1, leaving n=4 (which is W for opponent), or take 4, leaving n=1 (which is W for opponent). So, both moves leave opponent in W positions, so n=5 is L.\n- If n=6: current player can take 1, leaving n=5 (L for opponent), so current player can win by taking 1. So, W.\n- If n=7: current player can take 1, leaving n=6 (W for opponent), or take 4, leaving n=3 (W for opponent). So, both moves leave opponent in W positions, so n=7 is L.\n- If n=8: current player can take 1, leaving n=7 (L for opponent), so current player can win by taking 1. So, W.\n- If n=9: current player can take 1, leaving n=8 (W for opponent), or take 4, leaving n=5 (L for opponent). So, current player can win by taking 4. So, W.\n- If n=10: current player can take 1, leaving n=9 (W for opponent), or take 4, leaving n=6 (W for opponent). So, both moves leave opponent in W positions, so n=10 is L.\n \nHmm, so let\u0027s list out the W and L positions from n=1 to n=10:\n\nn: 1 2 3 4 5 6 7 8 9 10\n\nW/L: W L W W W L W L W L\n\nWait, let me check that again:\n\nn=1: W\n\nn=2: current player takes 1, opponent gets n=1 (W for opponent), so current player loses. So, n=2: L\n\nn=3: current player takes 1, opponent gets n=2 (L for opponent), so current player can win. So, n=3: W\n\nn=4: current player can take 1 or 4. Taking 1 leaves n=3 (W for opponent), taking 4 leaves n=0 (L for opponent). So, current player can win by taking 4. So, n=4: W\n\nn=5: current player takes 1, opponent gets n=4 (W); takes 4, opponent gets n=1 (W). So, both moves leave opponent in W positions, so n=5: L\n\nn=6: current player takes 1, opponent gets n=5 (L). So, current player can win. So, n=6: W\n\nn=7: current player takes 1, opponent gets n=6 (W); takes 4, opponent gets n=3 (W). So, both moves leave opponent in W positions, so n=7: L\n\nn=8: current player takes 1, opponent gets n=7 (L). So, current player can win. So, n=8: W\n\nn=9: current player takes 1, opponent gets n=8 (W); takes 4, opponent gets n=5 (L). So, current player can win by taking 4. So, n=9: W\n\nn=10: current player takes 1, opponent gets n=9 (W); takes 4, opponent gets n=6 (W). So, both moves leave opponent in W positions, so n=10: L\n\nSo, the pattern seems to be L at n=2,5,7,10,... Let me write down the L positions so far: 2,5,7,10,...\n\nWait, let me check n=11:\n\nn=11: current player can take 1, leaving n=10 (L for opponent), so current player can win by taking 1. So, n=11: W\n\nn=12: current player can take 1, leaving n=11 (W); take 4, leaving n=8 (W). So, both moves leave opponent in W positions. So, n=12: L\n\nn=13: current player can take 1, leaving n=12 (L). So, current player can win by taking 1. So, n=13: W\n\nn=14: current player can take 1, leaving n=13 (W); take 4, leaving n=10 (L). So, current player can win by taking 4. So, n=14: W\n\nn=15: current player can take 1, leaving n=14 (W); take 4, leaving n=11 (W). So, both moves leave opponent in W positions. So, n=15: L\n\nn=16: current player can take 1, leaving n=15 (L). So, current player can win by taking 1. So, n=16: W\n\nn=17: current player can take 1, leaving n=16 (W); take 4, leaving n=13 (W). So, both moves leave opponent in W positions. So, n=17: L\n\nn=18: current player can take 1, leaving n=17 (L). So, current player can win by taking 1. So, n=18: W\n\nn=19: current player can take 1, leaving n=18 (W); take 4, leaving n=15 (L). So, current player can win by taking 4. So, n=19: W\n\nn=20: current player can take 1, leaving n=19 (W); take 4, leaving n=16 (W). Both moves leave opponent in W positions. So, n=20: L\n\nSo, the L positions so far are: 2,5,7,10,12,15,17,20,...\n\nLet me see if I can find a pattern here. Let\u0027s list them:\n\n2,5,7,10,12,15,17,20,22,25,27,30,...\n\nWait, let\u0027s compute the differences between consecutive L positions:\n\n5-2=3\n\n7-5=2\n\n10-7=3\n\n12-10=2\n\n15-12=3\n\n17-15=2\n\n20-17=3\n\n22-20=2\n\n25-22=3\n\n27-25=2\n\n30-27=3\n\nSo, the pattern is alternating differences of 3 and 2. So, starting at 2, then adding 3, then 2, then 3, etc.\n\nSo, the L positions are numbers congruent to 2 mod 5 and 0 mod 5? Wait, let me check:\n\n2 mod 5 is 2\n\n5 mod 5 is 0\n\n7 mod 5 is 2\n\n10 mod 5 is 0\n\n12 mod 5 is 2\n\n15 mod 5 is 0\n\n17 mod 5 is 2\n\n20 mod 5 is 0\n\n22 mod 5 is 2\n\n25 mod 5 is 0\n\n27 mod 5 is 2\n\n30 mod 5 is 0\n\nYes, so it alternates between numbers that are 0 mod 5 and 2 mod 5. So, the losing positions are numbers that are 0 or 2 mod 5.\n\nWait, hold on. Let me check:\n\nn=2: 2 mod 5=2, yes.\n\nn=5: 5 mod5=0.\n\nn=7: 7 mod5=2.\n\nn=10:10 mod5=0.\n\nn=12:12 mod5=2.\n\nn=15:15 mod5=0.\n\nn=17:17 mod5=2.\n\nn=20:20 mod5=0.\n\nYes, so the losing positions are numbers that are 0 or 2 mod5.\n\nSo, for n\u22610 mod5 or n\u22612 mod5, it\u0027s a losing position.\n\nWait, but when n=5, which is 0 mod5, and n=2, which is 2 mod5, both are losing positions.\n\nBut let\u0027s test n=25 and n=27, which should be losing positions.\n\nn=25: If it\u0027s 0 mod5, then it\u0027s L.\n\nn=26: current player can take 1, leaving n=25 (L). So, n=26: W\n\nn=27: current player can take 1, leaving n=26 (W); take 4, leaving n=23. Wait, let me check n=23.\n\nWait, maybe I should check n=23.\n\nWait, maybe I need to test further.\n\nWait, perhaps my initial analysis was correct up to n=20, but n=23 may not follow the same pattern. Let me compute n=21 to n=25.\n\nWait, n=21:\n\nn=21: current player can take 1, leaving n=20 (L). So, n=21: W\n\nn=22: current player can take 1, leaving n=21 (W); take 4, leaving n=18 (W). So, both moves leave opponent in W positions, so n=22: L\n\nn=23: current player can take 1, leaving n=22 (L). So, n=23: W\n\nn=24: current player can take 1, leaving n=23 (W); take 4, leaving n=20 (L). So, current player can win by taking 4, leaving n=20 (L). So, n=24: W\n\nn=25: current player can take 1, leaving n=24 (W); take 4, leaving n=21 (W). So, both moves leave opponent in W positions, so n=25: L\n\nSo, n=25: L, which is 0 mod5.\n\nSimilarly, n=26: W, as before.\n\nn=27: current player can take 1, leaving n=26 (W); take 4, leaving n=23 (W). So, both moves leave opponent in W positions, so n=27: L\n\nYes, so n=27 is L, which is 2 mod5.\n\nSo, the pattern seems consistent.\n\nTherefore, the losing positions are n \u22610 mod5 and n\u22612 mod5.\n\nTherefore, for Bob to have a winning strategy, n must be a losing position for Alice, i.e., n \u22610 or 2 mod5.\n\nWait, but hold on. The way the problem is stated, Alice goes first. So, if n is a losing position for Alice, meaning that no matter what she does, Bob can win. So, yes, Bob can win when n is a losing position for Alice, which are n\u22610 or 2 mod5.\n\nTherefore, the number of positive integers n \u22642024 where Bob can force a win is equal to the number of n \u22642024 such that n \u22610 or 2 mod5.\n\nSo, now, we need to compute the count of numbers \u22642024 that are \u22610 or 2 mod5.\n\nLet me compute how many numbers \u22642024 are \u22610 mod5 and how many are \u22612 mod5.\n\nFirst, for numbers \u22610 mod5:\n\nThe smallest is 5, then 10,15,... up to the largest \u22642024.\n\nWhat\u0027s the largest multiple of 5 \u22642024? 2024 \u00f75=404.8, so 404\u00d75=2020.\n\nSo, numbers are 5,10,...,2020.\n\nNumber of terms: 2020 /5=404.\n\nSimilarly, numbers \u22612 mod5:\n\nThese are 2,7,12,..., up to the largest \u22642024.\n\nLet me find the largest number \u22612 mod5 \u22642024.\n\n2024 \u00f75=404.8, so 404\u00d75=2020, then 2020+2=2022.\n\nSo, the sequence is 2,7,12,...,2022.\n\nNumber of terms: Let me compute how many terms.\n\nFirst term:2, last term:2022, common difference:5.\n\nNumber of terms= ((2022 -2)/5)+1= (2020/5)+1=404+1=405.\n\nTherefore, total numbers \u22610 mod5:404, \u22612 mod5:405.\n\nTotal numbers:404 +405=809.\n\nBut wait, n must be positive integers, so n starts at 1. However, n=0 is not considered as per the problem.\n\nWait, in our analysis, n=0 is losing, but n starts at 1. So, n=0 isn\u0027t counted. So, the numbers \u22610 mod5 are from 5 to2020, which is 404 numbers, and numbers \u22612 mod5 are from2 to2022, which is405 numbers.\n\nTherefore, total numbers where Bob can win:404 +405=809.\n\nBut let me verify: 5\u00d7404=2020, correct. For 2 mod5, 2022=5\u00d7404 +2, so 405 numbers. So, yes, total is809.\n\nWait, but hold on. Let me think again.\n\nIf we count n from1 to2024, inclusive.\n\nThe number of integers \u22610 mod5 is floor(2024/5)=404 (since 5\u00d7404=2020).\n\nThe number of integers \u22612 mod5 is floor((2024 -2)/5) +1= floor(2022/5)+1=404 +1=405, as above.\n\nSo, total is404 +405=809.\n\nBut wait, let me check for n=1:1\u22611 mod5, which is a winning position, so Alice can win.\n\nn=2: losing position, Bob can win.\n\nn=3: winning, Alice can win.\n\nn=4: winning, Alice can win.\n\nn=5: losing, Bob can win.\n\nSo, the first few numbers where Bob can win are2,5,7,10,12,15,17,20,22,25,27,...\n\nWhich is exactly numbers \u22610 or2 mod5.\n\nSo, the count is809.\n\nBut hold on, wait. Let\u0027s make sure that for n=2022, which is a losing position, it\u0027s included since2022\u22642024.\n\nYes, 2022 is included.\n\nSimilarly, n=2025 would be the next multiple of5, but since2025\u003e2024, it\u0027s not included.\n\nSimilarly, n=2027 would be the next number \u22612 mod5, but2027\u003e2024, so not included.\n\nTherefore, total numbers where Bob can win is809.\n\nWait, but let me just compute 404 +405:\n\n404 +405=809.\n\nBut wait, 404 +405=809, correct.\n\nBut wait, the total number of integers from1 to2024 is2024.\n\nNumber of losing positions is809, meaning the remaining are winning positions.\n\nBut 2024 -809=1215.\n\nBut wait, 2024 divided by5 is404.8, so 404 numbers for each residue, but since2024 is not a multiple of5, some residues have one more number.\n\nWait, actually, in the range1 to2024, residues mod5:\n\nEach residue from0 to4 has either404 or405 numbers.\n\nSpecifically, residues0 and2 have404 and405?\n\nWait, no, let me compute.\n\nTotal numbers:2024.\n\nNumber per residue: floor(2024/5)=404 with remainder4.\n\nSo, residues1,2,3,4 will have405 numbers each, and residue0 has404.\n\nWait, no, actually, when you distribute2024 into5 residue classes, each class will havefloor(2024/5)=404, and the remainder is2024%5=2024 -5\u00d7404=2024-2020=4. So, residues1,2,3,4 will have1 more each, so405 each, and residue0 has404.\n\nTherefore, the number of numbers \u22610 mod5 is404, and numbers\u22611,2,3,4 mod5 are405 each.\n\nWait, but in our case, the losing positions are\u22610 and2 mod5.\n\nSo, number of losing positions=404 +405=809.\n\nHence, the count is809.\n\nTherefore, the number of positive integers n \u22642024 for which Bob can force a win is809.\n\nBut before finalizing, let me cross-verify with another approach.\n\nAnother way to think is that the losing positions are periodic with period5, repeating every5 numbers, with two losing positions in each period: positions2 and5, which are2 and0 mod5.\n\nSo, in each block of5 numbers, there are2 losing positions.\n\nTotal number of blocks in2024 numbers:2024//5=404 full blocks, each contributing2 losing positions:404\u00d72=808.\n\nThen, the remaining numbers after404\u00d75=2020 numbers are2021 to2024, which are4 numbers.\n\nIn these4 numbers, check if any are losing positions.\n\nThe losing positions are\u22610 or2 mod5.\n\n2021\u22611 mod5, 2022\u22612 mod5,2023\u22613 mod5,2024\u22614 mod5.\n\nSo, among2021-2024, only2022 is a losing position.\n\nTherefore, total losing positions=404\u00d72 +1=809.\n\nYes, same result.\n\nTherefore, I can be confident that the answer is809.\n\n**Final Answer**\nThe number of positive integers \\( n \\) less than or equal to 2024 for which Bob can force a win is \\(\\boxed{809}\\).\n\u003c/think\u003e\n\nTo determine the number of positive integers \\( n \\leq 2024 \\) for which Bob can force a win, we analyze the game where Alice and Bob take turns removing either 1 or 4 tokens from a stack. The player who removes the last token wins.\n\nWe define positions as winning (W) or losing (L) based on whether the current player can force a win. Through analysis, we find that the losing positions (where the current player cannot force a win) are those where \\( n \\equiv 0 \\pmod{5} \\) or \\( n \\equiv 2 \\pmod{5} \\).\n\nTo find the number of such losing positions \\( n \\leq 2024 \\):\n\n1. Calculate the number of integers \\( n \\leq 2024 \\) such that \\( n \\equiv 0 \\pmod{5} \\):\n - The largest multiple of 5 less than or equal to 2024 is 2020.\n - The number of such integers is \\( \\frac{2020}{5} = 404 \\).\n\n2. Calculate the number of integers \\( n \\leq 2024 \\) such that \\( n \\equiv 2 \\pmod{5} \\):\n - The largest such integer less than or equal to 2024 is 2022.\n - The number of such integers is \\( \\frac{2022 - 2}{5} + 1 = 405 \\).\n\nAdding these, the total number of losing positions is \\( 404 + 405 = 809 \\).\n\nThus, the number of positive integers \\( n \\leq 2024 \\) for which Bob can force a win is \\(\\boxed{809}\\).",
"letter": null,
"link": "https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_3",
"main_category": "Discrete Mathematics",
"model_name": "deepseek-ai/DeepSeek-R1-Distill-Qwen-7B",
"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/r1-7b-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/r1-7b-distil")
```

</details>

size_categories: 样本规模类别：样本数 (n < 1000)
dataset_info:
数据集信息:
features:
- 名称：题目ID（problem_id），数据类型：字符串
- 名称：链接（link），数据类型：字符串
- 名称：题目（problem），数据类型：字符串
- 名称：解析（solution），数据类型：字符串
- 名称：选项字母（letter），数据类型：空值
- 名称：答案（answer），数据类型：浮点型
- 名称：年份（year），数据类型：整型
- 名称：赛季/届次（season），数据类型：空值
- 名称：赛事（competition），数据类型：字符串
- 名称：版本（version），数据类型：字符串
- 名称：题目编号（problem_number），数据类型：整型
- 名称：类别（category），数据类型：字符串
- 名称：主类别（main_category），数据类型：字符串
- 名称：子类别（subcategory），数据类型：字符串
- 名称：生成方式（generation），数据类型：字符串
- 名称：distilabel元数据（distilabel_metadata），数据类型：结构体，包含：
- 原始输入文本生成项0（raw_input_text_generation_0）：列表类型，内部元素为结构体，包含：
- 内容（content）：字符串类型
- 角色（role）：字符串类型
- 原始输出文本生成项0（raw_output_text_generation_0）：字符串类型
- 生成统计项0（statistics_text_generation_0）：结构体，包含：
- 输入Token数（input_tokens）：整型
- 输出Token数（output_tokens）：整型
- 名称：模型名称（model_name），数据类型：字符串
splits:
- 名称：训练集（train），字节数：48515507，样本数：951
download_size: 20101605
dataset_size: 48515507
configs:
- 配置名称（config_name）：default，数据文件：
- 划分（split）：train，路径：data/train-*
tags:
- 合成数据（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="基于Distilabel构建" width="200" height="32"/>
</a>
</p>

# r1-7b-distil 数据集卡片

本数据集通过[distilabel](https://distilabel.argilla.io/)构建。

## 数据集概述
本数据集包含一个`pipeline.yaml`文件，可通过`distilabel`命令行工具复现构建该数据集的流程：
console
distilabel pipeline run --config "https://huggingface.co/datasets/AlgorithmicOps/r1-7b-distil/raw/main/pipeline.yaml"

或查看该配置的详细信息：
console
distilabel pipeline info --config "https://huggingface.co/datasets/AlgorithmicOps/r1-7b-distil/raw/main/pipeline.yaml"

## 数据集结构
每种配置下的样本结构如下：

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

json
{
"答案": 809.0,
"类别": "离散数学：组合博弈论",
"赛事": "AIME",
"distilabel元数据": {
"原始输入文本生成项0": [
{
"内容": "请逐步推导该问题，并将最终答案置于\boxed{}中：
爱丽丝与鲍勃玩如下游戏：桌上有一堆共(n)枚代币，两人轮流操作，爱丽丝先手。每回合玩家可移除1枚或4枚代币，移除最后一枚代币的玩家获胜。求小于等于2024的正整数(n)的个数，使得存在鲍勃的必胜策略，即无论爱丽丝如何操作，鲍勃均可获胜。",
"角色": "用户"
}
],
"原始输出文本生成项0": "好的，我们来逐步分析这个博弈论问题。首先，我们定义两种位置：若当前玩家可通过合理操作确保获胜，则称该位置为必胜位置（记为W）；若无论当前玩家如何操作，对手均可获胜，则称该位置为必败位置（记为L）。

首先考虑小数值的(n)：
- (n=1)：当前玩家直接拿走1枚代币获胜，为W位置。
- (n=2)：当前玩家只能拿走1枚，剩余1枚由对手拿走获胜，当前玩家失败，为L位置。
- (n=3)：当前玩家拿走1枚，剩余2枚（对手处于L位置），因此当前玩家可获胜，为W位置。
- (n=4)：当前玩家直接拿走4枚获胜，为W位置。
- (n=5)：当前玩家拿走1枚剩余4枚（对手W位置），拿走4枚剩余1枚（对手W位置），无论如何操作对手均可获胜，为L位置。

继续推导更多数值的情况，我们可以发现必败位置的规律：当(n equiv 0 pmod{5})或(n equiv 2 pmod{5})时，当前玩家处于必败位置。这是因为若当前玩家拿走1枚，对手可拿走4枚使剩余代币数仍为5的倍数或余2；若当前玩家拿走4枚，对手可拿走1枚达到同样效果，最终会将对手逼入必败位置。

由于爱丽丝是先手，因此当(n)为必败位置时，鲍勃即可获胜。接下来我们计算满足条件的(n)的个数：
1. 满足(n equiv 0 pmod{5})的数：最小为5，最大为2020（(5 imes 404 = 2020)），共404个。
2. 满足(n equiv 2 pmod{5})的数：最小为2，最大为2022（(5 imes 404 + 2 = 2022)），共(frac{2022 - 2}{5} + 1 = 405)个。

总计满足条件的数为(404 + 405 = 809)，因此最终答案为(\boxed{809}).",
"生成统计项0": {
"输入Token数": 129,
"输出Token数": 5678
}
},
"生成方式": "好的，我们来逐步分析这个博弈论问题。首先，我们定义两种位置：若当前玩家可通过合理操作确保获胜，则称该位置为必胜位置（记为W）；若无论当前玩家如何操作，对手均可获胜，则称该位置为必败位置（记为L）。

首先考虑小数值的(n)：
- (n=1)：当前玩家直接拿走1枚代币获胜，为W位置。
- (n=2)：当前玩家只能拿走1枚，剩余1枚由对手拿走获胜，当前玩家失败，为L位置。
- (n=3)：当前玩家拿走1枚，剩余2枚（对手处于L位置），因此当前玩家可获胜，为W位置。
- (n=4)：当前玩家直接拿走4枚获胜，为W位置。
- (n=5)：当前玩家拿走1枚剩余4枚（对手W位置），拿走4枚剩余1枚（对手W位置），无论如何操作对手均可获胜，为L位置。

继续推导更多数值的情况，我们可以发现必败位置的规律：当(n equiv 0 pmod{5})或(n equiv 2 pmod{5})时，当前玩家处于必败位置。这是因为若当前玩家拿走1枚，对手可拿走4枚使剩余代币数仍为5的倍数或余2；若当前玩家拿走4枚，对手可拿走1枚达到同样效果，最终会将对手逼入必败位置。

由于爱丽丝是先手，因此当(n)为必败位置时，鲍勃即可获胜。接下来我们计算满足条件的(n)的个数：
1. 满足(n equiv 0 pmod{5})的数：最小为5，最大为2020（(5 imes 404 = 2020)），共404个。
2. 满足(n equiv 2 pmod{5})的数：最小为2，最大为2022（(5 imes 404 + 2 = 2022)），共(frac{2022 - 2}{5} + 1 = 405)个。

总计满足条件的数为(404 + 405 = 809)，因此最终答案为(\boxed{809}).",
"选项字母": null,
"链接": "https://artofproblemsolving.com/wiki/index.php/2024_AIME_I_Problems/Problem_3",
"主类别": "离散数学",
"模型名称": "deepseek-ai/DeepSeek-R1-Distill-Qwen-7B",
"题目": "爱丽丝和鲍勃玩如下游戏。桌上有一堆共(n)枚代币。两人轮流操作，爱丽丝先手。每回合玩家可以从堆中移除1枚或4枚代币。移除最后一枚代币的玩家获胜。求小于等于2024的正整数(n)的个数，使得存在鲍勃的策略，确保无论爱丽丝如何操作，鲍勃都能获胜。",
"题目ID": "f098258808bd480c688d1a313d743992",
"题目编号": 3,
"赛季/届次": null,
"解析": "我们首先通过枚举小数值的情况分析博弈位置：
- 当(n=1)时，爱丽丝直接拿走1枚获胜；
- 当(n=2)时，爱丽丝只能拿走1枚，剩余1枚由鲍勃拿走，鲍勃获胜；
- 当(n=3)时，爱丽丝拿走1枚，鲍勃拿走1枚，爱丽丝拿走最后1枚获胜；
- 当(n=4)时，爱丽丝直接拿走4枚获胜；
- 当(n=5)时，无论爱丽丝拿走1枚还是4枚，鲍勃均可拿走剩余代币获胜。

由此可归纳规律：若(n)为5的倍数或除以5余2，则当前玩家处于必败位置。由于爱丽丝先手，当(n)为必败位置时，鲍勃可获胜。

接下来计算满足条件的(n)的个数：
1. 满足(n equiv 0 pmod{5})的数：最大为2020，共(2020/5=404)个；
2. 满足(n equiv 2 pmod{5})的数：最大为2022，共((2022-2)/5 +1=405)个。

总计(404+405=809)，故答案为(\boxed{809}).",
"子类别": "组合博弈论",
"版本": "I",
"年份": 2024
}

该子集可通过如下代码加载：
python
from datasets import load_dataset

ds = load_dataset("AlgorithmicOps/r1-7b-distil", "default")

或直接通过以下方式加载，由于仅存在一个名为default的配置：
python
from datasets import load_dataset

ds = load_dataset("AlgorithmicOps/r1-7b-distil")