2002 AIME I 第 8 题

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8.

求最小整数 kk,使得下列条件
(1)(1) a1a_1,a2a_2,a3a_3,…\ldots 是一个非递减的正整数数列
(2)(2) an=an−1+an−2a_n = a_{n-1} + a_{n-2} 对所有 n>2n \gt 2 都成立
(3)(3) a9=ka_9 = k
能同时被不止一个数列满足。
Find the smallest integer kk for which the conditions
(1)(1) a1,a_1, a2,a_2, a3,a_3, …\ldots is a nondecreasing sequence of positive integers
(2)(2) an=an−1+an−2a_n = a_{n-1} + a_{n-2} for all n>2n \gt 2
(3)(3) a9=ka_9 = k
are satisfied by more than one sequence.
答案:748
知识点:递推斐波那契数列丢番图方程
难度评级:2650
小提示:
反复使用递推式,把目标项表示成前两项:a9=13a1+21a2a_9 = 13a_1 + 21a_2。
Iterate the recurrence to express the target in the first two terms: a9=13a1+21a2a_9 = 13a_1 + 21a_2
大提示:
若 13x+21y=13u+21v13x + 21y = 13u + 21v 且 x<ux \lt u,则 u−xu - x 是 2121 的倍数,迫使 uu 和 vv 至少为 2222。
If 13x+21y=13u+21v13x + 21y = 13u + 21v with x<u,x \lt u, then u−xu - x is a multiple of 21,21, forcing uu and vv to be at least 2222
解答:
反复使用递推式得 a9=13a1+21a2a_9 = 13a_1 + 21a_2,且数列非递减当且仅当 0<a1≤a20 \lt a_1 \le a_2 (之后的项会自动满足)。因此我们需要找最小的 kk,使得 13x+21y=k13x + 21y = k 有两个满足 0<x≤y0 \lt x \le y 的解。
假设 13x+21y=13u+21v13x + 21y = 13u + 21v 且 x<ux \lt u。则 13(u−x)=21(y−v)13(u - x) = 21(y - v),所以 u−xu - x 是 2121 的正倍数。因此 u≥x+21≥22u \ge x + 21 \ge 22,又因为 u≤vu \le v,也有 v≥22v \ge 22,从而 k=13u+21v≥34⋅22=748k = 13u + 21v \ge 34 \cdot 22 = 748。
反过来,k=748k = 748 确实可行:(x,y)=(1,35)(x, y) = (1, 35) 和 (22,22)(22, 22) 分别给出数列 11、3535、3636、7171、107107、178178、285285、463463、748748,以及 2222、2222、4444、6666、110110、176176、286286、462462、748748。所以答案是 k=748k = 748。
Iterating the recurrence gives a9=13a1+21a2,a_9 = 13a_1 + 21a_2, and the sequence is nondecreasing exactly when 0<a1≤a20 \lt a_1 \le a_2 (all later terms then take care of themselves). So we need the smallest kk for which 13x+21y=k13x + 21y = k has two solutions with 0<x≤y.0 \lt x \le y.
Suppose 13x+21y=13u+21v13x + 21y = 13u + 21v with x<u.x \lt u. Then 13(u−x)=21(y−v),13(u - x) = 21(y - v), so u−xu - x is a positive multiple of 21.21. Hence u≥x+21≥22,u \ge x + 21 \ge 22, and since u≤v,u \le v, also v≥22,v \ge 22, giving k=13u+21v≥34⋅22=748.k = 13u + 21v \ge 34 \cdot 22 = 748.
Conversely k=748k = 748 works: (x,y)=(1,35)(x, y) = (1, 35) and (22,22)(22, 22) give the sequences 1,1, 35,35, 36,36, 71,71, 107,107, 178,178, 285,285, 463,463, 748748 and 22,22, 22,22, 44,44, 66,66, 110,110, 176,176, 286,286, 462,462, 748.748. The answer is k=748.k = 748.
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