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

Revision as of 19:28, 20 March 2017

Problem

Let $m$ be a positive integer, and let $a_0, a_1,\ldots,a_m$ be a sequence of reals 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 $0 < k < m$ , 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 $m = \boxed{889}$ , our answer.

@@ Line 1: / Line 1: @@
 == Problem ==
-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 reals 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>
-''Note: Clearly, the stipulation that the sequence is composed of integers is a minor oversight, as the term <math>a_2 </math>, for example, is obviously not integral.''
 == Solution ==

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 19:28, 20 March 2017

Problem

Solution

See also