# 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 reals 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 $m.$

## Solution 1

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.

Note: In order for $a_{m} = 0$ we need $a_{m-1}a_{m}=3$ simply by the recursion definition.

## Solution 2

Plugging in $k = m-1$ to the given relation, we get $0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}$. Inspecting the value of $a_{k}a_{k+1}$ for small values of $k$, we see that $a_{k}a_{k+1} = 37\cdot 72 - 3k$. Setting the RHS of this equation equal to $3$, we find that $m$ must be $\boxed{889}$.

~ anellipticcurveoverq

## Video solution

 2005 AIME II (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions