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

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== See Also ==
 
== See Also ==
  
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*[[2005 AIME II Problems/Problem 10| Previous problem]]
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*[[2005 AIME II Problems/Problem 12| Next problem]]
 
* [[2005 AIME II Problems]]
 
* [[2005 AIME II Problems]]
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 20:07, 7 September 2006

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 $m.$

Solution

For $\displaystyle 0 < k < m$, we have

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

Thus the product $a_{k}a_{k+1}$ is a monovariant: it decreases by 3 each time $\displaystyle k$ increases by 1. Since for $\displaystyle k = 0$ 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 $\displaystyle m = 889$, our answer.

See Also

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