Automated MNLP evaluation report (2026-05-13)

EVAL_REPORT.md

@@ -2,7 +2,7 @@
 
 - **Model repo:** [`cs-552-2026-centralesupechec/math_model`](https://huggingface.co/cs-552-2026-centralesupechec/math_model)
 - **Owner(s):** group **centralesupechec**
-- **Generated at:** 2026-05-13T04:17:52+00:00 (UTC)
+- **Generated at:** 2026-05-13T23:30:17+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 | 0.1700 | ok |
+| Math | 0.1400 | ok |
 | Knowledge | — | not run |
 | Multilingual | — | not run |
 | Safety | — | not run |
@@ -24,82 +24,62 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
 **Correct** (1 shown)
 
-- **reference**: `2`
-- **overall** (3/8 completions correct)
-- **extracted** (✓): `2`
+- **reference**: `.35625`
+- **overall** (8/8 completions correct)
+- **extracted** (✓): `0.35625`
 - **completion**:
 
     ```text
     <think>
-    Okay, so I need to solve the equation $3^{2x} + 19 = 10^x$. Let me think about how to approach this. 
+    Okay, so I need to figure out how to write 57 divided by 160 as a terminating decimal. Hmm, let me recall what a terminating decimal is. From what I remember, a decimal terminates if the denominator, after simplifying the fraction, has no prime factors other than 2 and 5. So, if the denominator is only 2s and 5s, then the decimal will terminate. If there are other prime factors, like 3, 7,
     
-    First, let me rewrite the equation to see if I can simplify it or recognize any patterns. The left side has $3^{2x}$, which is the same as $(3^2)^x = 9^x$. So, the equation becomes $9^x + 19 = 10^x$. 
+    ... [truncated 823 chars] ...
     
-    Hmm, so now it's $9^x + 19 = 10^x$. I don't see an obvious way to combine these terms,
+    er confirmation.
     
-    ... [truncated 821 chars] ...
+    Therefore, I can be confident that the terminating decimal form of 57/160 is 0.35625.
     
-    9^x + 19 - 10^x
-    $$
+    **Final Answer**
+    The terminating decimal form of $\frac{57}{160}$ is \boxed{0.35625}.
+    </think>
     
-    We seek the value(s) of $x$ for which $f(x) = 0$.
+    To determine the terminating decimal form of the fraction $\frac{57}{160}$, we begin by analyzing its denominator.
     
     ---
     
-    ### Step 2: Test Integer Values
+    ### Step 1: Check for Terminating Decimal
     
-    Let’s test some integer values of $x$:
+    A fraction will have a **terminating decimal** if, after simplifying the fraction, the **denominator (in lowest terms)** has **only the prime factors 2 and 5**. 
     
-    - **At $x = 2$:**
+    - The denominator is $160 = 2^5 \times 5^1$
+    - The numerator $57$ is not divisible by any prime other than 3 and 19, and it is already in lowest terms with the denominator.
     
-    $$
-    f(2) = 9^2 + 19 - 10^2 = 81 + 19 - 100 = 100 - 100 = 0
-    $$
-    
-    So, $x = 2$ is a solution.
+    Since the denominator only contains the prime factors $2$ and $5$, the decimal representation of $\frac{57}{160}$ is **terminating**.
     
     ---
     
-    ### Step 3: Analyze the Behavior of the Function
+    ### Step 2: Convert to Decimal
     
-    Define the function:
+    We now compute the decimal value of $\frac{57}{160}$ by performing the division:
     
     $$
-    f(x) = 9^x + 19 - 10^x
+    \frac{57}{160} = 0.35625
     $$
     
-    As $x \to \infty$, the exponential term $10^x$ dominates, so $f(x) \to -\infty$.
-    
-    As $x \to -\infty$, both $9^x$ and $10^x$ approach 0, so $f(x) \to 19$.
-    
-    Now, compute the derivative:
+    This can be verified by multiplying the result by the original denominator:
     
     $$
-    f'(x) = \ln(9) \cdot 9^x - \ln(10) \cdot 10^x
+    0.35625 \times 160 = 57
     $$
     
-    This derivative is always negative for all $x$, since:
-    
-    - $\ln(9) \approx 2.1972$
-    - $\ln(10) \approx 2.3026$
-    - $9^x$ grows slower than $10^x$, and the coefficient of $9^x$ is smaller than that of $10^x$
-    
-    Thus, $f'(x) < 0$ for all $x$, meaning the function is strictly decreasing.
-    
-    ---
-    
-    ### Step 4: Conclude the Number of Solutions
-    
-    Since $f(x)$ is strictly decreasing and continuous, and it goes from $19$ at $-\infty$ to $-\infty$ at $+\infty$, it can cross the x-axis **exactly once**.
-    
-    We already found that $f(2) = 0$, so this is the **only** solution.
+    Thus, the decimal representation is correct and terminates.
     
     ---
     
     ### Final Answer
     
     $$
-    \boxed{2}
+    \boxed{0.35625}
     $$
     ```
 
@@ -112,19 +92,27 @@ _Prompts are intentionally omitted to avoid revealing benchmark contents. For mu
 
     ```text
     <think>
-    Okay, so I need to figure out the greatest positive integer that can't be expressed as a sum of elements from the set C_n, where C_n is defined as {2^n - 2^i | 0 ≤ i < n}. And n is a positive integer greater than or equal to 2. Hmm, let's start by understanding what exactly the set C_n looks like.
+    Okay, so I need to figure out the greatest positive integer that can't be expressed as a sum of elements from the set C_n, where C_n is defined as {2^n - 2^i : i is an integer and 0 ≤ i < n}. And n is a positive integer greater than or equal to 2. Alright, let me start by understanding the problem step by step.
+    
+    First, let me try to write down what C_n looks like. For example, if n is 2, t
+    
+    ... [truncated 823 chars] ...
+    
+    ers starting from 1:
+    
+    1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, etc.
     
-    First, let's take a specific example. Let's say n = 2. Then C_2 would be {2^2 - 2^0, 2^2 - 2
+    We saw that 1-7 can't be formed. 8 can. 9-11 can't. 12 can. 13 can't. 14 can. 15 can. 16 can. 17 can't. 18 can't. 19 can't. 20 can. 21 can't. 22: 8 + 14. Yes. 23: 8 + 15. Yes. 24: 12 + 12. Yes. 25: 12 + 13? No. 15 + 10? No. 8 + 17? No. Wait, 25: 8 + 8 + 9? No. 12 + 13? No. 14 + 11? No. 15 + 10? No. So 25 can't be formed? Wait, 25 is 8*3 + 1? No. 12*2 + 1? No. 14 + 11? No. 15 + 10? No. So 25 can't be formed. Hmm. So maybe there's a lot of numbers that can't be formed. But how do we find the largest one?
     
-    ... [truncated 822 chars] ...
+    Alternatively, maybe there's a pattern here. For n=2, answer is 1. For n=3, answer is 9. For n=4, maybe answer is something else?
     
-    3 = 24, okay. So 21 is the largest non-representable number for n=4? But that seems high. Maybe I made a mistake here.
+    Wait, maybe I need to find a general pattern here. Let me think.
     
-    Wait, let's check 21 again. How can we get 21? The elements are 8, 12, 14, 15. Let me think. 15 + 8 = 23. 14 + 8 = 22. 12 + 12 = 24. 12 + 14 = 26. 15 + 12 = 27. 15 + 14 = 29. 8 + 14 = 22. 8 + 15 = 23. 12 + 15 = 27. So 21 is not in the list. So 21 can't be formed. 22 is 8 + 14. 23 is 8 + 15. 24 is 8 + 8 + 8. 25: 12 + 13? No. 14 + 11? No. 15 + 10? No. 8 + 12 + 5? No. 8 + 14 + 3? No. So 25 can't be formed? Wait, but 25 is 8 + 8 + 9? No. 12 + 12 + 1? No. 14 + 11? No. 15 + 10? No. So 25 can't be formed. So 25 is not representable.
+    Looking at n=2: C_n = {2, 3}. The largest non-representable is 1.
     
-    Wait, so 21, 25, 26? Wait, 26 is 12 + 14. So 26 is okay. So 25 is not representable. Then 26 is okay. So 25 is not representable. Then 27 is 15 + 12. So 27 is okay. So 25 is not representable. So 25 is larger than 21. So maybe the largest non-representable number is 25?
+    n=3: C_n = {4, 6, 7}. The largest non-representable is 9.
     
-    Wait, but how do we get 25? Let me think again. 8, 12, 14, 15. Let me try combinations:
+    n=4: C_n = {8, 12, 14, 15}. Maybe the answer is 23? Or something else?
     
-    25 - 15 = 10. 10 can't be formed. 25 - 14 = 11. 11 can't be formed. 25 - 12 = 13. 13 can't be formed. 25 - 8 = 17. 17 can't be formed. So none of these combinations work. So 25 can't be
+    Wait, let me try to think of the general case. Let me try to think of the elements of C_n. For general n, C_n is {2^n - 2^i | 0 ≤ i < n}. So, for example, when n=2, 2^2 - 2^0 = 4 - 1 = 3, and 2^2 -
     ```