Difference between revisions of "2005 AIME II Problems/Problem 11"
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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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+ | 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== |
Latest revision as of 14:08, 7 October 2020
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.
Note: In order for we need simply by the recursion definition.
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
Video solution
https://www.youtube.com/watch?v=JfxNr7lv7iQ
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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