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math_model/EVAL_REPORT.md
ModelHub XC c3e312fc28 初始化项目,由ModelHub XC社区提供模型
Model: cs-552-2026-centralesupechec/math_model
Source: Original Platform
2026-06-13 10:10:18 +08:00

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# Automated MNLP evaluation report
- **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)
- **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._
## Summary
| Benchmark | Accuracy | Status |
|---|---:|---|
| Math | 0.1700 | 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**: `2`
- **overall** (3/8 completions correct)
- **extracted** (✓): `2`
- **completion**:
```text
<think>
Okay, so I need to solve the equation $3^{2x} + 19 = 10^x$. Let me think about how to approach this.
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 $9^x + 19 = 10^x$.
Hmm, so now it's $9^x + 19 = 10^x$. I don't see an obvious way to combine these terms,
... [truncated 821 chars] ...
9^x + 19 - 10^x
$$
We seek the value(s) of $x$ for which $f(x) = 0$.
---
### Step 2: Test Integer Values
Lets test some integer values of $x$:
- **At $x = 2$:**
$$
f(2) = 9^2 + 19 - 10^2 = 81 + 19 - 100 = 100 - 100 = 0
$$
So, $x = 2$ is a solution.
---
### Step 3: Analyze the Behavior of the Function
Define the function:
$$
f(x) = 9^x + 19 - 10^x
$$
As $x \to \infty$, the exponential term $10^x$ dominates, so $f(x) \to -\infty$.
As $x \to -\infty$, both $9^x$ and $10^x$ approach 0, so $f(x) \to 19$.
Now, compute the derivative:
$$
f'(x) = \ln(9) \cdot 9^x - \ln(10) \cdot 10^x
$$
This derivative is always negative for all $x$, since:
- $\ln(9) \approx 2.1972$
- $\ln(10) \approx 2.3026$
- $9^x$ grows slower than $10^x$, and the coefficient of $9^x$ is smaller than that of $10^x$
Thus, $f'(x) < 0$ for all $x$, meaning the function is strictly decreasing.
---
### Step 4: Conclude the Number of Solutions
Since $f(x)$ is strictly decreasing and continuous, and it goes from $19$ at $-\infty$ to $-\infty$ at $+\infty$, it can cross the x-axis **exactly once**.
We already found that $f(2) = 0$, so this is the **only** solution.
---
### Final Answer
$$
\boxed{2}
$$
```
**Incorrect** (1 shown)
- **reference**: `$(n-2)2^n +1$`
- **overall** (0/8 completions correct)
- **extracted** (✗): `<no answer>`
- **completion**:
```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.
First, let's take a specific example. Let's say n = 2. Then C_2 would be {2^2 - 2^0, 2^2 - 2
... [truncated 822 chars] ...
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, 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.
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?
Wait, but how do we get 25? Let me think again. 8, 12, 14, 15. Let me try combinations:
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
```