Problem Analysis #4

28.73s
6,203 toks
Problem Statement

Suppose the following limit is given:

limn[0,n](1+20569nx)69ne(67x)dx=pq,\lim_{n\to\infty} \int_{[0,n]} \left(1+\frac{205}{69n}x\right)^{-69n} e^{(-\frac{6}{7}x)}\,dx=\frac{p}{q},

where pp and qq are positive integers which are relatively prime. Find p2qp^2q.

oindent

Ground Truth
70609
Execution Trace
Reasoning Process

[THOUGHT] The limit is (\lim_{n\to\infty} \int_0^n \left(1 + \frac{205}{69n} x \right)^{-69n} e^{-\frac{6}{7}x} , dx).

Pointwise, as (n \to \infty), (\left(1 + \frac{205}{69n} x \right)^{-69n} \to e^{-205 x}).

Thus, the integrand converges pointwise to (e^{-205 x} e^{-\frac{6}{7} x} = e^{-\left(205 + \frac{6}{7}\right) x}).

The limit of the integral is (\int_0^\infty e^{-\left(205 + \frac{6}{7}\right) x} , dx = \frac{1}{205 + \frac{6}{7}} = \frac{1}{\frac{1435 + 6}{7}} = \frac{7}{1441}).

Verify (\gcd(7, 1441) = 1) and compute (p^2 q = 7^2 \times 1441 = 49 \times 1441). [/THOUGHT] [PYTHON] import math print(math.gcd(7, 1441)) print(49 * 1441) [/PYTHON]

Tool Output
[PYTHON OUTPUT] 1 70609 None [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The calculation confirms (\gcd(7,1441)=1) and (7^2 \times 1441 = 70609). The limit is (\frac{7}{1441}), so (p=7), (q=1441), and (p^2 q = 70609). [/THOUGHT] \boxed{70609}

Explanation

The reasoning begins by analyzing the pointwise limit of the integrand as (n \to \infty). For fixed (x), the term (\left(1 + \frac{205}{69n} x \right)^{-69n}) converges to (e^{-205 x}), since (\left(1 + \frac{a}{n}\right)^{n b} \to e^{a b}) with (a = \frac{205}{69} x) and (b = -69), yielding (a b = -205 x).

Thus, the integrand (\left(1 + \frac{205}{69n} x \right)^{-69n} e^{-\frac{6}{7} x}) converges pointwise to (e^{-205 x} e^{-\frac{6}{7} x} = e^{-\left(205 + \frac{6}{7}\right) x}).

The upper limit of integration is (n), which approaches infinity, so the limit of the integral equals (\int_0^\infty e^{-\left(205 + \frac{6}{7}\right) x} , dx).

Compute the exponent: (205 + \frac{6}{7} = \frac{205 \cdot 7 + 6}{7} = \frac{1435 + 6}{7} = \frac{1441}{7}).

The integral evaluates to (\int_0^\infty e^{-\frac{1441}{7} x} , dx = \frac{7}{1441}).

Here, (\frac{7}{1441}) is in lowest terms, as verified by (\gcd(7, 1441) = 1).

Thus, (p = 7) and (q = 1441), so (p^2 q = 7^2 \cdot 1441 = 49 \cdot 1441 = 70609).

Justification for interchanging limit and integral relies on the integrand being non-negative and the pointwise convergence to an integrable function, allowing application of the monotone convergence theorem.

Final Model Output
70609

Comparative Analysis (P4) (Average across runs)

Avg Time Taken

Deepseek-R1-distill-Qwen-32B (high)
140.9
Deepseek-V3.2
123.2
Gemini-3-Pro-Preview
90.8
GPT-oss-20B (high)
60.4
Solar-Pro-2 (31B)(high)
60.2
Grok-4.1-fast
38.4
HCX-007(high)
31.5
Gemma-3-27B
31.4
Qwen3-30B-A3B-2507
28.9
Llama-VARCO-8B-Instruct
21.3
GPT-5.1 (high)
19.0
Claude-Opus-4.5
16.4
A.X-4.0 (72B)
16.0
EXAONE-4.0.1-32B (high)
3.7

Avg Token Usage

Gemini-3-Pro-Preview
11241.0
Deepseek-V3.2
9874.7
Solar-Pro-2 (31B)(high)
9831.3
GPT-oss-20B (high)
9757.0
Grok-4.1-fast
7671.7
Qwen3-30B-A3B-2507
6266.7
HCX-007(high)
4190.3
Deepseek-R1-distill-Qwen-32B (high)
3984.0
Claude-Opus-4.5
3858.3
A.X-4.0 (72B)
3485.3
EXAONE-4.0.1-32B (high)
3238.3
GPT-5.1 (high)
3067.7
Llama-VARCO-8B-Instruct
2088.3
Gemma-3-27B
1705.7