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Model: cs-552-2026-4neurons/math_model
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# Automated MNLP evaluation report
- **Model repo:** [`cs-552-2026-4neurons/math_model`](https://huggingface.co/cs-552-2026-4neurons/math_model)
- **Owner(s):** group **4neurons**
- **Generated at:** 2026-06-08T04:40:54+00:00 (UTC)
- **Pipeline:** [mnlp-project-ci](https://github.com/eric11eca/mnlp-project-ci)
_This PR is opened automatically by the course CI. It is **non-blocking** — you do not need to merge it. The next nightly run will refresh this file._
## Evaluated checkpoint
- **Commit:** [`9b57663`](https://huggingface.co/cs-552-2026-4neurons/math_model/commit/9b57663282076a91efe6f5eec880675c011f6f0e)
- **Message:** Upload checkpoint-200 math model with patched math chat template
- **Committed:** 2026-06-06T19:42:53+00:00
## Summary
| Benchmark | Accuracy | Status |
|---|---:|---|
| Math | 0.4400 | ok |
| Knowledge | — | not run |
| Multilingual | — | not run |
| Safety | — | not run |
## Sample completions
_Prompts are intentionally omitted to avoid revealing benchmark contents. For multi-completion problems, only one completion is shown per sample._
### Math
**Correct** (1 shown)
- **reference**: `305`
- **overall** (2/8 completions correct)
- **extracted** (✓): `305`
- **completion**:
```text
<think>
Okay, let's try to tackle this problem step by step. So, we need to find all positive integers n between 99 and 206 such that the numbers 1 to 2n can be partitioned into n pairs where the differences within each pair are exactly 1, 2, ..., n. Then, among all such n, find the smallest (m) and the largest (M) and compute m + M.
First, let's understand the problem. We have to partition the s
... [truncated 822 chars] ...
into \( n \) pairs with differences exactly \( 1, 2, \ldots, n \). We then find the least and greatest elements \( m \) and \( M \) of \( S \) and compute \( m + M \).
1. **Understanding the Problem**:
- We need to partition the set \(\{1, 2, \ldots, 2n\}\) into \( n \) pairs such that each pair has a unique difference from 1 to \( n \).
- Each difference must be used exactly once.
2. **Key Observations**:
- For each difference \( k \) from 1 to \( n \), there must be exactly one pair of numbers in \(\{1, 2, \ldots, 2n\}\) with difference \( k \).
- This is equivalent to finding a permutation where each element is paired with another element such that the difference is unique.
3. **Existence of Such Pairings**:
- Research and analysis show that such pairings (called difference permutations) exist for all \( n \geq 1 \).
- The smallest \( n \) in the given range is 99, and the largest is 206.
4. **Conclusion**:
- The smallest \( n \) is 99, and the largest \( n \) is 206.
- Therefore, the least element \( m \) is 99, the greatest element \( M \) is 206, and their sum is \( 99 + 206 = 305 \).
\[
\boxed{305}
\]
Therefore, the final answer is \boxed{305}.
```
**Incorrect** (1 shown)
- **reference**: `7200`
- **overall** (0/8 completions correct)
- **extracted** (✗): `1441200`
- **completion**:
```text
<think>
Okay, let's try to tackle this problem step by step. So, we need to find the number of polynomials P with integer coefficients that satisfy three conditions. Let's break them down one by one.
First, the polynomial P has degree at most 3, so it's of the form P(x) = ax³ + bx² + cx + d, where a, b, c, d are integers. The coefficients are between 0 and 2012, inclusive. So each of the coeffici
... [truncated 822 chars] ...
r all \( x \). Simplifying, we get \( 2b \equiv 0 \pmod{2013} \) and \( 2d \equiv 0 \pmod{2013} \). Since 2013 is odd, this implies \( b \equiv 0 \pmod{2013} \) and \( d \equiv 0 \pmod{2013} \). Thus, \( b = 0 \) and \( d = 0 \), simplifying \( P(x) \) to \( ax^3 + cx \).
Next, we need \( P(x) = ax^3 + cx \) to have distinct residues modulo 2013 for \( x = 0, 1, \ldots, 2012 \). This requires \( ax^3 + cx \) to be a permutation polynomial modulo 2013.
For \( P(x) \) to be a permutation polynomial modulo 2013, \( a \) and \( c \) must be invertible modulo 2013. The number of invertible elements modulo 2013 is given by Euler's totient function \( \phi(2013) \). Since \( 2013 = 3 \times 11 \times 61 \), we have \( \phi(2013) = 2 \times 10 \times 60 = 1200 \).
There are two cases:
1. \( a = 0 \) and \( c \) is invertible. There are 1200 choices for \( c \).
2. \( a \neq 0 \) and both \( a \) and \( c \) are invertible. There are \( 1200 \times 1200 = 1,440,000 \) choices for \( a \) and \( c \).
Summing these cases, the total number of polynomials is \( 1200 + 1,440,000 = 1,441,200 \).
Thus, the final answer is \(\boxed{1441200}\).
Therefore, the final answer is \boxed{1441200}.
```

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---
library_name: transformers
tags: []
---
# Model Card for Model ID
<!-- Provide a quick summary of what the model is/does. -->
## Model Details
### Model Description
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This is the model card of a 🤗 transformers model that has been pushed on the Hub. This model card has been automatically generated.
- **Developed by:** [More Information Needed]
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## Uses
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### Direct Use
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### Out-of-Scope Use
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Users (both direct and downstream) should be made aware of the risks, biases and limitations of the model. More information needed for further recommendations.
## How to Get Started with the Model
Use the code below to get started with the model.
[More Information Needed]
## Training Details
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### Training Procedure
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#### Preprocessing [optional]
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#### Training Hyperparameters
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#### Speeds, Sizes, Times [optional]
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### Testing Data, Factors & Metrics
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Carbon emissions can be estimated using the [Machine Learning Impact calculator](https://mlco2.github.io/impact#compute) presented in [Lacoste et al. (2019)](https://arxiv.org/abs/1910.09700).
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{%- set enable_thinking = true %}
{%- set default_system_prompt = "You are a rigorous competition-math solver. Solve problems from prealgebra, algebra, number theory, combinatorics, probability, geometry, precalculus, and calculus.\n\nUse only information stated in the problem. Do not invent missing facts, hidden diagram properties, extra assumptions, or numerical values. \n\nWork carefully and verify each important step. Prefer exact symbolic reasoning over decimal approximations. Keep track of definitions, constraints, and edge cases.\n\nFor geometry, justify angle, length, arc, parallel, cyclic, and similarity claims before using them.\nFor counting and probability, avoid double-counting and check whether cases are disjoint and exhaustive.\nFor algebra, check domains and discard extraneous solutions.\n\nIf there are multiple plausible approaches, choose the most reliable one and complete it. If you are uncertain, give the best-supported answer from your derivation rather than making up unsupported facts.\n\nYour response may include concise reasoning, but the final answer must appear at the very end in LaTeX boxed format: \\\\boxed{answer}. Do not write anything after the final \\\\boxed{answer}.\nExamples: If the answer is 42, write \\\\boxed{42}, if the answer is p-q, write \\\\boxed{p - q}.\nFor polynomials, the answer should first display the higher order and be reduced as much as possible. Example \\\\boxed{x^3 - 300x^2 + 30000x - 999900} instead of \\\\boxed{- 999900 + x^3 - 300x^2 + 30000x } (incorrect order) or \\\\boxed{x^3 - 300x^2 + 30000x - 1000000 + 100} (not reduced)" %}
{%- if messages[0]['role'] != 'system' %}
{%- set messages = [{'role': 'system', 'content': default_system_prompt}] + messages %}
{%- endif %}
{%- if tools %}
{{- '<|im_start|>system\n' }}
{%- if messages[0].role == 'system' %}
{{- messages[0].content + '\n\n' }}
{%- endif %}
{{- "# Tools\n\nYou may call one or more functions to assist with the user query.\n\nYou are provided with function signatures within <tools></tools> XML tags:\n<tools>" }}
{%- for tool in tools %}
{{- "\n" }}
{{- tool | tojson }}
{%- endfor %}
{{- "\n</tools>\n\nFor each function call, return a json object with function name and arguments within <tool_call></tool_call> XML tags:\n<tool_call>\n{\"name\": <function-name>, \"arguments\": <args-json-object>}\n</tool_call><|im_end|>\n" }}
{%- else %}
{%- if messages[0].role == 'system' %}
{{- '<|im_start|>system\n' + messages[0].content + '<|im_end|>\n' }}
{%- endif %}
{%- endif %}
{%- set ns = namespace(multi_step_tool=true, last_query_index=messages|length - 1) %}
{%- for message in messages[::-1] %}
{%- set index = (messages|length - 1) - loop.index0 %}
{%- if ns.multi_step_tool and message.role == "user" and message.content is string and not(message.content.startswith('<tool_response>') and message.content.endswith('</tool_response>')) %}
{%- set ns.multi_step_tool = false %}
{%- set ns.last_query_index = index %}
{%- endif %}
{%- endfor %}
{%- for message in messages %}
{%- if message.content is string %}
{%- set content = message.content %}
{%- else %}
{%- set content = '' %}
{%- endif %}
{%- if (message.role == "user") or (message.role == "system" and not loop.first) %}
{{- '<|im_start|>' + message.role + '\n' + content + '<|im_end|>' + '\n' }}
{%- elif message.role == "assistant" %}
{%- set reasoning_content = '' %}
{%- if message.reasoning_content is string %}
{%- set reasoning_content = message.reasoning_content %}
{%- else %}
{%- if '</think>' in content %}
{%- set reasoning_content = content.split('</think>')[0].rstrip('\n').split('<think>')[-1].lstrip('\n') %}
{%- set content = content.split('</think>')[-1].lstrip('\n') %}
{%- endif %}
{%- endif %}
{%- if loop.index0 > ns.last_query_index %}
{%- if loop.last or (not loop.last and reasoning_content) %}
{{- '<|im_start|>' + message.role + '\n<think>\n' + reasoning_content.strip('\n') + '\n</think>\n\n' + content.lstrip('\n') }}
{%- else %}
{{- '<|im_start|>' + message.role + '\n' + content }}
{%- endif %}
{%- else %}
{{- '<|im_start|>' + message.role + '\n' + content }}
{%- endif %}
{%- if message.tool_calls %}
{%- for tool_call in message.tool_calls %}
{%- if (loop.first and content) or (not loop.first) %}
{{- '\n' }}
{%- endif %}
{%- if tool_call.function %}
{%- set tool_call = tool_call.function %}
{%- endif %}
{{- '<tool_call>\n{"name": "' }}
{{- tool_call.name }}
{{- '", "arguments": ' }}
{%- if tool_call.arguments is string %}
{{- tool_call.arguments }}
{%- else %}
{{- tool_call.arguments | tojson }}
{%- endif %}
{{- '}\n</tool_call>' }}
{%- endfor %}
{%- endif %}
{{- '<|im_end|>\n' }}
{%- elif message.role == "tool" %}
{%- if loop.first or (messages[loop.index0 - 1].role != "tool") %}
{{- '<|im_start|>user' }}
{%- endif %}
{{- '\n<tool_response>\n' }}
{{- content }}
{{- '\n</tool_response>' }}
{%- if loop.last or (messages[loop.index0 + 1].role != "tool") %}
{{- '<|im_end|>\n' }}
{%- endif %}
{%- endif %}
{%- endfor %}
{%- if add_generation_prompt %}
{{- '<|im_start|>assistant\n' }}
{%- if enable_thinking is defined and enable_thinking is false %}
{{- '<think>\n\n</think>\n\n' }}
{%- endif %}
{%- endif %}

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{%- set enable_thinking = true %}
{%- if tools %}
{{- '<|im_start|>system\n' }}
{%- if messages[0].role == 'system' %}
{{- messages[0].content + '\n\n' }}
{%- endif %}
{{- "# Tools\n\nYou may call one or more functions to assist with the user query.\n\nYou are provided with function signatures within <tools></tools> XML tags:\n<tools>" }}
{%- for tool in tools %}
{{- "\n" }}
{{- tool | tojson }}
{%- endfor %}
{{- "\n</tools>\n\nFor each function call, return a json object with function name and arguments within <tool_call></tool_call> XML tags:\n<tool_call>\n{\"name\": <function-name>, \"arguments\": <args-json-object>}\n</tool_call><|im_end|>\n" }}
{%- else %}
{%- if messages[0].role == 'system' %}
{{- '<|im_start|>system\n' + messages[0].content + '<|im_end|>\n' }}
{%- endif %}
{%- endif %}
{%- set ns = namespace(multi_step_tool=true, last_query_index=messages|length - 1) %}
{%- for message in messages[::-1] %}
{%- set index = (messages|length - 1) - loop.index0 %}
{%- if ns.multi_step_tool and message.role == "user" and message.content is string and not(message.content.startswith('<tool_response>') and message.content.endswith('</tool_response>')) %}
{%- set ns.multi_step_tool = false %}
{%- set ns.last_query_index = index %}
{%- endif %}
{%- endfor %}
{%- for message in messages %}
{%- if message.content is string %}
{%- set content = message.content %}
{%- else %}
{%- set content = '' %}
{%- endif %}
{%- if (message.role == "user") or (message.role == "system" and not loop.first) %}
{{- '<|im_start|>' + message.role + '\n' + content + '<|im_end|>' + '\n' }}
{%- elif message.role == "assistant" %}
{%- set reasoning_content = '' %}
{%- if message.reasoning_content is string %}
{%- set reasoning_content = message.reasoning_content %}
{%- else %}
{%- if '</think>' in content %}
{%- set reasoning_content = content.split('</think>')[0].rstrip('\n').split('<think>')[-1].lstrip('\n') %}
{%- set content = content.split('</think>')[-1].lstrip('\n') %}
{%- endif %}
{%- endif %}
{%- if loop.index0 > ns.last_query_index %}
{%- if loop.last or (not loop.last and reasoning_content) %}
{{- '<|im_start|>' + message.role + '\n<think>\n' + reasoning_content.strip('\n') + '\n</think>\n\n' + content.lstrip('\n') }}
{%- else %}
{{- '<|im_start|>' + message.role + '\n' + content }}
{%- endif %}
{%- else %}
{{- '<|im_start|>' + message.role + '\n' + content }}
{%- endif %}
{%- if message.tool_calls %}
{%- for tool_call in message.tool_calls %}
{%- if (loop.first and content) or (not loop.first) %}
{{- '\n' }}
{%- endif %}
{%- if tool_call.function %}
{%- set tool_call = tool_call.function %}
{%- endif %}
{{- '<tool_call>\n{"name": "' }}
{{- tool_call.name }}
{{- '", "arguments": ' }}
{%- if tool_call.arguments is string %}
{{- tool_call.arguments }}
{%- else %}
{{- tool_call.arguments | tojson }}
{%- endif %}
{{- '}\n</tool_call>' }}
{%- endfor %}
{%- endif %}
{{- '<|im_end|>\n' }}
{%- elif message.role == "tool" %}
{%- if loop.first or (messages[loop.index0 - 1].role != "tool") %}
{{- '<|im_start|>user' }}
{%- endif %}
{{- '\n<tool_response>\n' }}
{{- content }}
{{- '\n</tool_response>' }}
{%- if loop.last or (messages[loop.index0 + 1].role != "tool") %}
{{- '<|im_end|>\n' }}
{%- endif %}
{%- endif %}
{%- endfor %}
{%- if add_generation_prompt %}
{{- '<|im_start|>assistant\n' }}
{%- if enable_thinking is defined and enable_thinking is false %}
{{- '<think>\n\n</think>\n\n' }}
{%- endif %}
{%- endif %}

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{
"architectures": [
"Qwen3ForCausalLM"
],
"attention_bias": false,
"attention_dropout": 0.0,
"bos_token_id": 151643,
"dtype": "bfloat16",
"eos_token_id": 151645,
"head_dim": 128,
"hidden_act": "silu",
"hidden_size": 2048,
"initializer_range": 0.02,
"intermediate_size": 6144,
"layer_types": [
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention",
"full_attention"
],
"max_position_embeddings": 40960,
"max_window_layers": 28,
"model_type": "qwen3",
"num_attention_heads": 16,
"num_hidden_layers": 28,
"num_key_value_heads": 8,
"pad_token_id": null,
"rms_norm_eps": 1e-06,
"rope_parameters": {
"rope_theta": 1000000,
"rope_type": "default"
},
"sliding_window": null,
"tie_word_embeddings": true,
"transformers_version": "5.7.0",
"use_cache": true,
"use_sliding_window": false,
"vocab_size": 151936
}

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You are a helpful math assistant. Solve the problem, show the reasoning, and end with the final answer in LaTeX \boxed{answer} format.

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{
"bos_token_id": 151643,
"do_sample": true,
"eos_token_id": [
151645,
151643
],
"pad_token_id": 151643,
"temperature": 0.7,
"top_k": 20,
"top_p": 0.9,
"transformers_version": "5.7.0"
}

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{
"adapter_dir": "/scratch/checkpoints/qwen3-1.7b-competition-simple-sft-lora",
"base_model_name_or_path": "/scratch/hf_cache/hub/models--Qwen--Qwen3-1.7B/snapshots/70d244cc86ccca08cf5af4e1e306ecf908b1ad5e",
"cache_dir": "/scratch/hf_cache",
"default_system_prompt_source": "/scratch/checkpoints/qwen3-1.7b-competition-simple-sft-merged/simple_sft_args.json",
"injected_default_system_prompt": true,
"max_shard_size": "4GB",
"safe_serialization": true,
"torch_dtype": "bfloat16"
}

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version https://git-lfs.github.com/spec/v1
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size 3441185608

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{
"attn_implementation": null,
"bf16": null,
"cache_dir": "/scratch/hf_cache",
"data_file": "/scratch/hf_cache/competition_math_train_sft",
"dataset_config": null,
"dataset_name": null,
"dry_run": false,
"filter_overlong": true,
"fp16": null,
"gradient_accumulation_steps": 8,
"gradient_checkpointing": true,
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