Automated MNLP evaluation report (2026-06-11)

EVAL_REPORT.md

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+# Automated MNLP evaluation report
+
+- **Model repo:** [`cs-552-2026-busybees/math_model`](https://huggingface.co/cs-552-2026-busybees/math_model)
+- **Owner(s):** group **busybees**
+- **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:** [`2f98a7f`](https://huggingface.co/cs-552-2026-busybees/math_model/commit/2f98a7f296cf22e3b8ed58d9a10030746e6edfbc)
+- **Message:** Upload folder using huggingface_hub
+- **Committed:** 2026-06-10T21:54:11+00:00
+
+## Summary
+
+| Benchmark | Accuracy | Status |
+|---|---:|---|
+| Math | 0.5200 | 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**: `70`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `70`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to find all integer bases b greater than 9 where the base-b number 17 divides the base-b number 97. Then sum all those bases. Hmm, let's start by understanding what the problem is asking.
+    
+    First, in base b, the number 17 is written as 1*b^1 + 7*b^0, right? Similarly, 97 in base b is 9*b^1 + 7*b^0. So, when we say that the base-b number 17 divides 97, we mean that when we co
+    
+    ... [truncated 823 chars] ...
+    
+    use the Euclidean algorithm:
+         \[
+         \frac{9b + 7}{b + 7} = 9 - \frac{56}{b + 7}
+         \]
+       - For this to be an integer, \( b + 7 \) must be a divisor of 56.
+    
+    4. **Identify the divisors of 56:**
+       - The divisors of 56 are: \( 1, 2, 4, 7, 8, 14, 28, 56 \).
+    
+    5. **Determine valid values for \( b \):**
+       - Since \( b > 9 \), we need \( b + 7 > 16 \).
+       - The divisors of 56 greater than 16 are: \( 28 \) and \( 56 \).
+       - Therefore, \( b + 7 = 28 \) or \( b + 7 = 56 \).
+       - Solving these, we get:
+         \[
+         b = 28 - 7 = 21 \quad \text{and} \quad b = 56 - 7 = 49
+         \]
+    
+    6. **Verify the solutions:**
+       - For \( b = 21 \):
+         \[
+         \frac{9 \cdot 21 + 7}{21 + 7} = \frac{189 + 7}{28} = \frac{196}{28} = 7
+         \]
+       - For \( b = 49 \):
+         \[
+         \frac{9 \cdot 49 + 7}{49 + 7} = \frac{441 + 7}{56} = \frac{448}{56} = 8
+         \]
+       - Both values of \( b \) satisfy the condition.
+    
+    7. **Sum the valid bases:**
+       - The valid bases are \( 21 \) and \( 49 \).
+       - The sum of these bases is:
+         \[
+         21 + 49 = 70
+         \]
+    
+    Thus, the sum of all integer values of the base \( b \) greater than 9 for which the base-\( b \) number 17 divides the base-\( b \) number 97 is \(\boxed{70}\).
+    ```
+
+**Incorrect** (1 shown)
+
+- **reference**: `60750, 854130, 854970`
+- **overall** (0/8 completions correct)
+- **extracted** (✗): `854130`
+- **completion**:
+
+    ```text
+    <think>
+    Okay, so I need to solve this problem where there's a function h from the natural numbers to natural numbers excluding 1. The condition given is that for every natural number n, h(n) + h(n+1) equals h(n+2) times h(n+3) minus 840. And the question is asking for every possible value that the sum of h(i) from i=1 to 2025 can take.
+    
+    Hmm, let's start by understanding the functional equation. Th
+    
+    ... [truncated 823 chars] ...
+    
+    \) is the only valid solution. Therefore, the constant solution is:
+    \[
+    h(n) = 30 \quad \text{for all} \quad n
+    \]
+    The sum from \( i = 1 \) to \( 2025 \) is:
+    \[
+    \sum_{i=1}^{2025} 30 = 2025 \times 30 = 60750
+    \]
+    
+    ### Step 2: Periodic Solution
+    Next, consider a periodic solution with period 2. Assume \( h(n) = 2 \) for even \( n \) and \( h(n) = 842 \) for odd \( n \). We need to verify if this satisfies the functional equation.
+    
+    For even \( n \):
+    \[
+    h(n) = 2, \quad h(n+1) = 842
+    \]
+    \[
+    h(n+2) = 842, \quad h(n+3) = 2
+    \]
+    Substitute into the equation:
+    \[
+    2 + 842 = 842 \times 2 - 840 \implies 844 = 1684 - 840 \implies 844 = 844
+    \]
+    This holds true. Similarly, for odd \( n \):
+    \[
+    h(n) = 842, \quad h(n+1) = 2
+    \]
+    \[
+    h(n+2) = 2, \quad h(n+3) = 842
+    \]
+    Substitute into the equation:
+    \[
+    842 + 2 = 2 \times 842 - 840 \implies 844 = 1684 - 840 \implies 844 = 844
+    \]
+    This also holds true. Therefore, the periodic solution with period 2 is valid.
+    
+    The sum from \( i = 1 \) to \( 2025 \) is:
+    \[
+    \sum_{i=1}^{2025} (2 + 842) = 1013 \times 2 + 1012 \times 842 = 2026 + 852104 = 854130
+    \]
+    
+    ### Conclusion
+    The possible sums for \( \sum_{i=1}^{2025} h(i) \) are:
+    \[
+    \boxed{60750} \quad \text{and} \quad \boxed{854130}
+    \]
+    ```