# 2002 AIME I Problems/Problem 8

## 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.

## 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 $k$.

## See also

 2002 AIME I (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 9 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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