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

Revision as of 20:00, 24 June 2014

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. 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.

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-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.
+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.  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 ==

2005 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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Difference between revisions of "2005 AIME II Problems/Problem 11"

Revision as of 20:00, 24 June 2014

Problem

Solution

See also