Problem Analysis #15

KT Mi:dm 2.0 Base Instruct · Run 1

Correct
1.24s
913 toks

API · no tools. This is the prompt and response recorded for this attempt; a full solution may not have been returned.

Problem Statement

Return a single integer only. The problem has been revised so that no final answer is an irrational number, a parameterized expression, a range, a proof, a drawing, a graph, a locus equation, or a region description.

The original limit value is

L=log⁡2−12.L=\log 2-\frac12.

To avoid a logarithmic final answer, compute the integer

eL+1/2.e^{L+1/2}.
Ground Truth
2
Recorded model response / solution
Reasoning Process

To compute eL+12e^{L + \frac{1}{2}} where L=log⁡2−12L = \log 2 - \frac{1}{2}, follow these steps:

  1. Substitute LL into the expression:

    eL+12=e(log⁡2−12)+12=elog⁡2.e^{L + \frac{1}{2}} = e^{\left(\log 2 - \frac{1}{2}\right) + \frac{1}{2}} = e^{\log 2}.

  2. Simplify using the property elog⁡a=ae^{\log a} = a:

    elog⁡2=2.e^{\log 2} = 2.

  3. Verify the result:

    • The original limit L=log⁡2−12L = \log 2 - \frac{1}{2} is approximately 0.193147180560.19314718056.
    • Adding 12\frac{1}{2} gives 0.19314718056+0.5=0.693147180560.19314718056 + 0.5 = 0.69314718056.
    • e0.69314718056≈2e^{0.69314718056} \approx 2, confirming the result.

Thus, the integer value is 2\boxed{2}.

Final Model Output
2

Comparative Analysis (P15) (Average across runs)

Averages use reported measurements only. Missing usage is not treated as zero.

Avg Time Taken

K-EXAONE-236B-A23B
34.9
DeepSeek V4 Pro
5.0
Gemini 3.5 Flash
4.3
GPT-5.5
4.1
Claude Opus 4.8
2.7
Solar Pro 3
1.0
KT Mi:dm 2.0 Base Instruct
0.9

Avg Token Usage

K-EXAONE-236B-A23B
2002.0
KT Mi:dm 2.0 Base Instruct
901.0
Gemini 3.5 Flash
795.0
DeepSeek V4 Pro
367.7
Solar Pro 3
299.0
Claude Opus 4.8
243.7
GPT-5.5
223.3
    Tokyo · Problem 15 · KT Mi:dm 2.0 Base Instruct | EntropyMath