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

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Let <math> \displaystyle m </math> be a positive integer, and let <math> a_0, a_1,\ldots,a_m </math> be a sequence of integers such that <math> \displaystyle a_0 = 37, a_1 = 72, a_m = 0, </math> and <math> a_{k+1} = a_{k-1} - \frac 3{a_k} </math> for <math> k = 1,2,\ldots, m-1. </math> Find <math> \displaystyle m. </math>
 
Let <math> \displaystyle m </math> be a positive integer, and let <math> a_0, a_1,\ldots,a_m </math> be a sequence of integers such that <math> \displaystyle a_0 = 37, a_1 = 72, a_m = 0, </math> and <math> a_{k+1} = a_{k-1} - \frac 3{a_k} </math> for <math> k = 1,2,\ldots, m-1. </math> Find <math> \displaystyle m. </math>
  
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''Note: Clearly, the stipulation that the sequence is composed of positive integers is a minor oversight, as the term <math>\displaystyle a_2 </math>, for example, is obviouly not integral.''
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''Note: Clearly, the stipulation that the sequence is composed of integers is a minor oversight, as the term <math>\displaystyle a_2 </math>, for example, is obviouly not integral.''
  
 
== Solution ==
 
== Solution ==

Revision as of 14:32, 2 March 2007

Problem

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

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

Solution

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

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

Thus the product $\displaystyle a_{k}a_{k+1}$ 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 $k = \frac{37 \cdot 72}{3} = 888$, $\displaystyle a_{k}a_{k+1}$ will be zero for the first time, which implies that $\displaystyle m = 889$, our answer.

See Also

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