Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-TopHaylin/math_model`](https://huggingface.co/cs-552-2026-TopHaylin/math_model)
+- **Owner(s):** group **TopHaylin**
+- **Generated at:** 2026-06-11T06:23:10+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:** [`220916a`](https://huggingface.co/cs-552-2026-TopHaylin/math_model/commit/220916ac14ae590b76c43e8fdb3107e08f884797)
+- **Message:** Upload folder using huggingface_hub
+- **Committed:** 2026-06-06T20:13:30+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.5000 | 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**: `244`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `244`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to find the number of positive integer divisors of $17017^{17}$ that are congruent to 5 modulo 12, and then find the remainder when that number is divided by 1000. Let me start by breaking down the problem step by step.
+    
+    First, let me factorize 17017. Hmm, 17017. Let me check if it's divisible by small primes. Let's try dividing by 7 first. 7*2431 is 17017? Let me calculate
+    
+    ... [truncated 822 chars] ...
+    
+    \equiv 5 \pmod{12} \Rightarrow (-1)^b \cdot 5 \equiv 5 \pmod{12}
+    $$
+    
+    Since $(-1)^b \cdot 5 \equiv 5 \pmod{12}$, we must have $b$ odd.
+    
+    - $a$ can be any of 9 odd values (1, 3, ..., 17)
+    - $d$ can be any of 9 even values (0, 2, ..., 16)
+    - $b$ can be any of 9 odd values (1, 3, ..., 17)
+    - $c$ can be any of 18 values (0, 1, ..., 17)
+    
+    So, the number of such divisors is:
+    
+    $$
+    9 \times 9 \times 9 \times 18 = 13122
+    $$
+    
+    #### Case 2: $a$ even, $d$ odd, $b$ even
+    
+    $$
+    7 \cdot (-1)^b \cdot 5 \equiv 5 \pmod{12} \Rightarrow (-1)^b \cdot 5 \equiv 5 \pmod{12}
+    $$
+    
+    Here, $b$ must be even.
+    
+    - $a$ can be any of 9 even values (0, 2, ..., 16)
+    - $d$ can be any of 9 odd values (1, 3, ..., 17)
+    - $b$ can be any of 9 even values (0, 2, ..., 16)
+    - $c$ can be any of 18 values (0, 1, ..., 17)
+    
+    So, the number of such divisors is also:
+    
+    $$
+    9 \times 9 \times 9 \times 18 = 13122
+    $$
+    
+    ---
+    
+    ### **Step 5: Total Count and Modulo Operation**
+    
+    Adding both cases, the total number of divisors congruent to $5 \mod 12$ is:
+    
+    $$
+    13122 + 13122 = 26244
+    $$
+    
+    Now, compute $26244 \mod 1000$:
+    
+    $$
+    26244 \equiv 244 \pmod{1000}
+    $$
+    
+    ---
+    
+    ### **Conclusion**
+    
+    $$
+    \text{The remainder when } N \text{ is divided by } 1000 \text{ is } \boxed{244}
+    $$
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `1, 13, 25, 325`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `25`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to find all positive integers n such that the number of positive divisors of n¹² is equal to n itself. Let me start by recalling some number theory concepts. 
+    
+    First, τ(m) is the divisor function, which counts the number of positive divisors of m. If I have a number n, then n¹² would be the number n raised to the 12th power. The number of divisors of a number can be determi
+    
+    ... [truncated 823 chars] ...
+    
+    ad \tau(1) = 1 \quad \Rightarrow \quad \tau(n^{12}) = n = 1
+    $$
+    
+    So, $ n = 1 $ is a solution.
+    
+    ---
+    
+    #### **Case 2: $ n = 25 $**
+    
+    $$
+    25 = 5^2 \quad \Rightarrow \quad \tau(25^{12}) = \tau(5^{24}) = 24 + 1 = 25
+    $$
+    
+    Thus, $ \tau(n^{12}) = 25 = n $, and $ n = 25 $ is a solution.
+    
+    ---
+    
+    #### **Case 3: $ n = 13 $**
+    
+    $$
+    13 = 13^1 \quad \Rightarrow \quad \tau(13^{12}) = 12 + 1 = 13
+    $$
+    
+    So, $ \tau(n^{12}) = 13 = n $, and $ n = 13 $ is a solution.
+    
+    ---
+    
+    #### **Case 4: $ n = 169 $**
+    
+    $$
+    169 = 13^2 \quad \Rightarrow \quad \tau(169^{12}) = \tau(13^{24}) = 24 + 1 = 25 \neq 169
+    $$
+    
+    So, $ \tau(n^{12}) \neq n $, and $ n = 169 $ is **not** a solution.
+    
+    ---
+    
+    ### Step 3: General Observations
+    
+    If $ n $ is not a prime power (i.e., $ n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} $), then $ \tau(n^{12}) $ becomes a product of terms like $ (12a_i + 1) $, and the resulting value must be equal to $ n $. This is only possible for small values of $ n $, as larger values tend to make $ \tau(n^{12}) $ grow much faster than $ n $ itself.
+    
+    ---
+    
+    ### Step 4: Conclusion
+    
+    We conclude that the only positive integers $ n $ satisfying $ \tau(n^{12}) = n $ are:
+    
+    $$
+    \boxed{1}, \quad \boxed{13}, \quad \text{and} \quad \boxed{25}
+    $$
+    ```