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

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== Solution ==
 
== Solution ==
  
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For <math>\displaystyle 0 < k < m</math>, we have
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For <math>0 < k < m</math>, we have
  
 
<center>
 
<center>
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<math>\displaystyle a_{k}a_{k+1} = a_{k-1}a_{k} - 3 </math>.
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<math>a_{k}a_{k+1} = a_{k-1}a_{k} - 3 </math>.
 
</center>
 
</center>
  
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Thus the product <math> \displaystyle 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> \displaystyle a_{k}a_{k+1}</math> will be zero for the first time, which implies that <math>\displaystyle m = 889</math>, our answer.
+
Thus the product <math>a_{k}a_{k+1}</math> is a [[monovariant]]: it decreases by 3 each time <math>k</math> increases by 1.  Since for <math>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>m = 889</math>, our answer.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Revision as of 22:22, 4 July 2013

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. Since 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 $m = 889$, our answer.

See also

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

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