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> | ||
| + | <math>\displaystyle a_{k}a_{k+1} = a_{k-1}a_{k} - 3 </math>. | ||
| + | </center> | ||
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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]] | + | |
| + | * [[2005 AIME II Problems]] | ||
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| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 17:11, 7 September 2006
Problem
Let 
be a positive integer, and let 
be a sequence of integers such that 
and 
for 
Find 
Solution
For 
, we have
.
Thus the product 
is a monovariant: it decreases by 3 each time 
increases by 1. Since for 
we have 
, so when 
, will be zero for the first time, which implies that 
, our answer.
