Problem Analysis #22

28.11s
6,127 toks
Problem Statement

Let f(x)f(x) be a continuous function defined on the interval [0,)[0,\infty) satisfying the following condition:When 2n11x<2n1,f(x)=2n1(x+1)22n2  +  2n11,\text{When } 2^{\,n-1}-1 \le x < 2^{\,n}-1,\quad f(x) = \sqrt{\,2^{\,n-1}(x+1) - 2^{\,2n-2}} \;+\; 2^{\,n-1} - 1,for n=1,2,3,n = 1, 2, 3, \ldots.Define a sequence {an}\{a_n\} by an=2n11a_n = 2^{\,n-1} - 1 for all natural numbers nn. Find the natural number nn such thatanan+1f(x)dx=14073748332216323.\int_{a_n}^{a_{n+1}} f(x)\, dx = \frac{1407374833221632}{3}.oindent

Ground Truth
25
Execution Trace
Reasoning Process

[THOUGHT] The integral from a_n to a_{n+1} simplifies to (5/3) k^2 - k where k = 2^{n-1}. Setting this equal to N/3 with N = 1407374833221632 gives 5k^2 - 3k = N. Solve the quadratic 5p^2 - 3p - N = 0 for p = [3 + sqrt(9 + 20N)] / 10, check if it's an integer power of 2, then n = log2(p) + 1. [/THOUGHT] [PYTHON] import math

N = 1407374833221632 D = 9 + 20 * N s = int(math.sqrt(D) + 0.5) is_square = (s * s == D) p_num = 3 + s if p_num % 10 == 0: p = p_num // 10 is_power_of_two = (p > 0) and (p & (p - 1) == 0) if is_power_of_two: exponent = p.bit_length() - 1 n = exponent + 1 print(n) else: print("Not power of 2") else: print("Not divisible by 10")

print("D:", D) print("sqrt approx:", s) print("is square:", is_square) print("p:", p_num // 10 if p_num % 10 == 0 else "not int") [/PYTHON]

Tool Output
[PYTHON OUTPUT] 25 D: 28147496664432649 sqrt approx: 167772157 is square: True p: 16777216 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python output confirms n=25. [/THOUGHT] \boxed{25}

Explanation

The reasoning begins by analyzing the integral ∫ from a_n to a_{n+1} of f(x) dx, where a_n = 2^{n-1} - 1 and f(x) is piecewise-defined. On the interval [a_n, a_{n+1}) = [2^{n-1} - 1, 2^n - 1), f(x) = √[2^{n-1}(x + 1) - 2^{2n-2}] + 2^{n-1} - 1. Substituting k = 2^{n-1} simplifies the integral to (5/3)k^2 - k.

Setting this equal to the given value 1407374833221632 / 3 yields the equation 5k^2 - 3k = 1407374833221632, or 5k^2 - 3k - N = 0 with N = 1407374833221632. Solving the quadratic equation gives k = [3 + √(9 + 20N)] / 10.

The Python code computes the discriminant D = 9 + 20N = 28147496664432649, verifies it is a perfect square (√D ≈ 167772157, and 167772157^2 = D), then calculates p_num = 3 + 167772157 = 167772160. Dividing by 10 yields p = 16777216, which is confirmed as a power of 2 (2^24, since 16777216 & 16777215 == 0).

Thus, k = p = 2^24, so n-1 = 24, giving n = 25. The output directly prints 25, confirming the solution.

Final Model Output
25

Comparative Analysis (P22) (Average across runs)

Avg Time Taken

Llama-VARCO-8B-Instruct
98.6
Deepseek-V3.2
92.3
Solar-Pro-2 (31B)(high)
84.2
HCX-007(high)
62.4
Gemini-3-Pro-Preview
40.7
A.X-4.0 (72B)
35.0
Claude-Opus-4.5
32.9
Grok-4.1-fast
28.1
GPT-5.1 (high)
25.6
EXAONE-4.0.1-32B (high)
5.2

Avg Token Usage

Solar-Pro-2 (31B)(high)
13592.0
Claude-Opus-4.5
9246.0
HCX-007(high)
6449.0
Deepseek-V3.2
6445.0
Grok-4.1-fast
6127.0
Llama-VARCO-8B-Instruct
5446.0
Gemini-3-Pro-Preview
4349.0
GPT-5.1 (high)
3937.0
EXAONE-4.0.1-32B (high)
3930.0
A.X-4.0 (72B)
2882.0