# Difference between revisions of "2005 AIME II Problems/Problem 11"

## Problem

Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of integers such that $a_0 = 37, a_1 = 72, a_m = 0,$ and $a_{k+1} = a_{k-1} - \frac 3{a_k}$ for $k = 1,2,\ldots, m-1.$ Find $m.$

Note: Clearly, the stipulation that the sequence is composed of integers is a minor oversight, as the term $a_2$, for example, is obviously not integral.

## Solution

For $0 < k < m$, we have

$a_{k}a_{k+1} = a_{k-1}a_{k} - 3$.

Thus the product $a_{k}a_{k+1}$ is a monovariant: it decreases by 3 each time $k$ increases by 1. For $k = 0$ we have $a_{k}a_{k+1} = 37\cdot 72$, so when $k = \frac{37 \cdot 72}{3} = 888$, $a_{k}a_{k+1}$ will be zero for the first time, which implies that $m = \boxed{889}$, our answer.