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

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== Problem ==
 
== Problem ==
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Let <math> 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> 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> m. </math>
 
Let <math> 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> 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> m. </math>
  
 
== Solution ==
 
== Solution ==
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For <math>\displaystyle 0 < k < m</math>, we have
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<center>
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<math>\displaystyle a_{k}a_{k+1} = a_{k-1}a_{k} - 3 </math>.
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Thus the product <math>a_{k}a_{k+1}</math> is a [[monovariant]]: it decreases by 3 each time <math>\displaystyle k</math> increases by 1.  Since for <math>\displaystyle k = 0</math> we have <math>a_{k}a_{k+1} = 37\cdot 72</math>, so when <math>k = \frac{37 \cdot 72}{3} = 888</math>, <math>a_{k}a_{k+1}</math> will be zero for the first time, which implies that <math>\displaystyle m = 889</math>, our answer.
  
 
== See Also ==
 
== See Also ==
*[[2005 AIME II Problems]]
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* [[2005 AIME II Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 17:11, 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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