1985 AHSME Problems/Problem 27
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Problem
Consider a sequence defined by: and in general
What is the smallest value of for which is an integer?
Solution
Firstly, we will show by induction that For the base case, we indeed have and for the inductive step, if our claim is true for , then which completes the proof.
We now rewrite our formula for as follows: and as is not a perfect power, we deduce that is an integer if and only if the exponent, , is itself an integer. By precisely the same argument, this reduces to being an integer, so the smallest possible (positive) value of is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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