Update Automated MNLP evaluation report (2026-06-01)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-claude-bots/math_model`](https://huggingface.co/cs-552-2026-claude-bots/math_model)
 - **Owner(s):** group **claude-bots**
-- **Generated at:** 2026-05-29T06:54:15+00:00 (UTC)
+- **Generated at:** 2026-06-01T06:08:54+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._
@@ -11,7 +11,7 @@ _This PR is opened automatically by the course CI. It is **non-blocking** — yo
 
 | Benchmark | Accuracy | Status |
 |---|---:|---|
-| Math | 0.2800 | ok |
+| Math | 0.2700 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,37 +24,30 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `4`
-- **overall** (3/8 completions correct)
-- **extracted** (✓): `4`
+- **reference**: `305`
+- **overall** (4/8 completions correct)
+- **extracted** (✓): `305`
 - **completion**:
 
     ```text
     <think>
-    The common ratio of the series is $12/a \div 4 = 3/a$.  The sum of an infinite geometric series is $\frac{4}{1 - 3/a} = \frac{4a}{a - 3}$.  We want this sum to be a perfect square, so we want $\frac{4a}{a - 3}$ to be a perfect square.  Since $a$ is a positive integer, $a - 3$ must be a positive divisor of $4a$.  The only way for $\frac{4a}{a - 3}$ to be a perfect square is if $a - 3$ divides $4a$ exactly once, so $a - 3$ must be a perfect square.  The smallest possible value of $a$ is $a - 3 = 1^2 = 1$, so $a = 4$.  The sum of the series is $\frac{4 \cdot 4}{4 - 3} = 16$, which is a perfect square.  Therefore, the least positive integer value of $a$ is $\boxed{4}$.
+    Condition (ii) implies that the differences between the numbers in each pair are exactly $1, 2, \ldots, n$. For example, if $n = 3$, the pairs could be $(1,2), (3,4), (5,6)$, with differences $1, 1, 1$, but this doesn't satisfy the condition. The correct pairing for $n = 3$ would be $(1,4), (2,5), (3,6)$, with differences $3, 3, 3$. This suggests that the pairs must be of the form $(k, k + n)$ for $k = 1, 2, \ldots, n$. Thus, the smallest $n$ is 99, and the largest is 206. Their sum is $99 + 206 = \boxed{305}$.
     </think>
     
-    \boxed{4}
+    \boxed{305}
     ```
 
 **Incorrect** (1 shown)
 
-- **reference**: `91/6`
+- **reference**: `(3,2,5)`
 - **overall** (0/8 completions correct)
-- **extracted** (✗): `\frac{13}{5}`
+- **extracted** (✗): `(3,1,2)`
 - **completion**:
 
     ```text
     <think>
-    Let $a$ be the number of red marbles Triton has, $b$ the number of green marbles, and $c$ the number of blue marbles. Initially, $a = b = c = 3$. At each turn, Triton has a $\frac{a}{6}$ chance of drawing a red marble, a $\frac{b}{6}$ chance of drawing a green marble, and a $\frac{c}{6}$ chance of drawing a blue marble. Thus, the expected number of turns until Triton has 6 marbles of one c
-    
-    ... [truncated 216 chars] ...
-    
-    \left(1 + \frac{b-2}{4} \cdot \left(1 + \dotsb \right)\right)\right) + \frac{c}{6} \cdot \left(1 + \frac{c-1}{5} \cdot \left(1 + \dotsb \right)\right).\]By symmetry, we can assume that $a \ge b \ge c$. Thus, $a \ge 6$ and $b = c = 3$. The expected number of turns until Triton has 6 red marbles and 3 green marbles is
-    \[1 + \frac{a}{6} \cdot \left(1 + \frac{a-1}{5} \cdot \left(1 + \frac{a-2}{4} \cdot \left(1 + \dotsb \right)\right)\right) + \frac{3}{6} \cdot \left(1 + \frac{2}{5} \cdot \left(1 + \dotsb \right)\right).\]By the formula for the expected value of a geometric distribution, the expected number of turns until Triton has 6 red marbles and 3 green marbles is
-    \[1 + \frac{a}{6} \cdot \frac{1}{1 - \frac{a-1}{5} \cdot \frac{1}{6}} + \frac{3}{6} \cdot \frac{1}{1 - \frac{2}{5} \cdot \frac{1}{6}}.\]If $a = 6$, then the expected number of turns is
-    \[1 + \frac{6}{6} \cdot \frac{1}{1 - \frac{5}{5} \cdot \frac{1}{6}} + \frac{3}{6} \cdot \frac{1}{1 - \frac{2}{5} \cdot \frac{1}{6}} = 1 + 1 + \frac{1}{1 - \frac{1}{6}} \cdot \frac{3}{6} = 2 + \frac{1}{\frac{5}{6}} \cdot \frac{3}{6} = 2 + \frac{6}{5} \cdot \frac{3}{6} = 2 + \frac{3}{5} = \boxed{\frac{13}{5}}.\]
+    Since $p$ is a prime greater than $2$, it is odd. The expression $m^{p-1} + n$ is a power of $p$, so $m^{p-1} + n = p^k$ for some integer $k$. Similarly, $n^{p-1} + m = p^l$ for some integer $l$. Since $m \le n$, $m^{p-1} \le n^{p-1}$, so $m^{p-1} + n \le n^{p-1} + n = p^k$. The only way this can happen is if $m^{p-1} + n = p$ and $n^{p-1} + m = p$, which implies $m = 1$ and $n = p - 1$. Thus, the only solution is $(p, m, n) = (3, 1, 2)$, since $p$ must be a prime greater than $2$.
     </think>
     
-    \boxed{\frac{13}{5}}
+    \boxed{(3,1,2)}
     ```