Difference between revisions of "2005 AIME II Problems/Problem 11"
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== Solution == | == Solution == | ||
− | For <math> | + | For <math>0 < k < m</math>, we have |
<center> | <center> | ||
− | <math> | + | <math>a_{k}a_{k+1} = a_{k-1}a_{k} - 3 </math>. |
</center> | </center> | ||
− | Thus the product <math> | + | 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. |
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 22:22, 4 July 2013
Problem
Let be a positive integer, and let be a sequence of integers such that and for Find
Note: Clearly, the stipulation that the sequence is composed of integers is a minor oversight, as the term , for example, is obviously not integral.
Solution
For , we have
.
Thus the product is a monovariant: it decreases by 3 each time increases by 1. Since for we have , so when , will be zero for the first time, which implies that , our answer.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.