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

Revision as of 14:08, 7 October 2020

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 1

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.

Note: In order for $a_{m} = 0$ we need $a_{m-2}a_{m-1}=3$ simply by the recursion definition.

Solution 2

Plugging in $k = m-1$ to the given relation, we get $0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}$ . Inspecting the value of $a_{k}a_{k+1}$ for small values of $k$ , we see that $a_{k}a_{k+1} = 37\cdot 72 - 3k$ . Setting the RHS of this equation equal to $3$ , we find that $m$ must be $\boxed{889}$ .

~ anellipticcurveoverq

Video solution

https://www.youtube.com/watch?v=JfxNr7lv7iQ

Revision as of 09:52, 8 September 2020 (view source) Jerry122805 (talk \| contribs) m (→‎Solution 1) ← Older edit		Revision as of 14:08, 7 October 2020 (view source) Jjmath123 (talk \| contribs) m (→‎Solution 1) Newer edit →
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−	Note: In order for <math>a_{m} = 0</math> we need <math>a_{m-1}a_{m}=3</math> simply by the recursion definition.	+	Note: In order for <math>a_{m} = 0</math> we need <math>a_{m-2}a_{m-1}=3</math> simply by the recursion definition.

	==Solution 2==		==Solution 2==

2005 AIME II (Problems • Answer Key • Resources)
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