Difference between revisions of "2005 AIME II Problems/Problem 11"
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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> | 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> | ||
− | == Solution == | + | ==Solution 1== |
For <math>0 < k < m</math>, we have | For <math>0 < k < m</math>, we have | ||
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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. | 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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+ | ==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 | ||
== See also == | == See also == |
Revision as of 11:48, 18 June 2020
Contents
Problem
Let be a positive integer, and let be a sequence of reals such that and for Find
Solution 1
For , we have
.
Thus the product is a monovariant: it decreases by 3 each time increases by 1. For we have , so when , will be zero for the first time, which implies that , our answer.
Solution 2
Plugging in to the given relation, we get . Inspecting the value of for small values of , we see that . Setting the RHS of this equation equal to , we find that must be .
~ anellipticcurveoverq
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
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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