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

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== Problem ==
 
== 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>
 
  
== Solution ==
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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 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>
  
== See Also ==
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==Solution 1==
*[[2005 AIME II Problems]]
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For <math>0 < k < m</math>, we have
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<center>
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<math>a_{k}a_{k+1} = a_{k-1}a_{k} - 3 </math>.
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</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>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 = \boxed{889}</math>, our answer.
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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.
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==Solution 2==
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Plugging in <math>k = m-1</math> to the given relation, we get <math>0 = a_{m-2} - \frac {3}{a_{m-1}} \implies{a_{m-2}a_{m-1} = 3}</math>. Inspecting the value of <math>a_{k}a_{k+1}</math> for small values of <math>k</math>, we see that <math>a_{k}a_{k+1} = 37\cdot 72 - 3k</math>. Setting the RHS of this equation equal to <math>3</math>, we find that <math>m</math> must be <math> \boxed{889}</math>.
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~ anellipticcurveoverq
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==Video solution==
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https://www.youtube.com/watch?v=JfxNr7lv7iQ
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== See also ==
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{{AIME box|year=2005|n=II|num-b=10|num-a=12}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest 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

See also

2005 AIME II (ProblemsAnswer KeyResources)
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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