AIME 2002 I · 第 8 题

AIME 2002 I — Problem 8

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Find the smallest integer $k$ for which the conditions

(1) $a_1,a_2,a_3\cdots$ is a nondecreasing sequence of positive integers

(2) $a_n=a_{n-1}+a_{n-2}$ for all $n>2$

(3) $a_9=k$

are satisfied by more than one sequence.

解析

Solution

From $(2)$ , $a_9=$ $a_8+a_7=2a_7+a_6=3a_6+2a_5=5a_5+3a_4=8a_4+5a_3=13a_3+8a_2=21a_2+13a_1$ $=k$

Suppose that $a_1=x_0$ is the smallest possible value for $a_1$ that yields a good sequence, and $a_2=y_0$ in this sequence. So, $13x_0+21y_0=k$ .

Since $\gcd(13,21)=1$ , the next smallest possible value for $a_1$ that yields a good sequence is $a_1=x_0+21$ . Then, $a_2=y_0-13$ .

By $(1)$ , $a_2 \ge a_1 \Rightarrow y_0-13 \ge x_0+21 \Rightarrow y_0 \ge x_0+34 \ge 35$ . So the smallest value of $k$ is attained when $(x_0,y_0)=(1,35)$ which yields $(a_1,a_2)=(1,35)$ or $(22,22)$ .

Thus, $k=13(1)+21(35)=\boxed{748}$ is the smallest possible value of $k$ .