Difference between revisions of "1981 IMO Problems/Problem 6"
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It follows that <math>\displaystyle f(4,1981) = 2^{2\cdot ^{ . \cdot 2}}</math> when there are 1984 2s, Q.E.D. | It follows that <math>\displaystyle f(4,1981) = 2^{2\cdot ^{ . \cdot 2}}</math> when there are 1984 2s, Q.E.D. | ||
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+ | {{alternate solutions}} | ||
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+ | == Resources == | ||
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+ | * [[1981 IMO Problems]] | ||
+ | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366648#p366648 AoPS/MathLinks Discussion] | ||
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+ | [[Category:Olympiad Algebra Problems]] |
Revision as of 17:36, 29 October 2006
Problem
The function satisfies
(1)
(2)
(3)
for all non-negative integers . Determine .
Solution
We observe that and that , so by induction, . Similarly, and , yielding .
We continue with ; ; ; and ; .
It follows that when there are 1984 2s, Q.E.D.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.