2009 AIME I Problems/Problem 7
Problem
The sequence satisfies
and
for
. Let
be the least integer greater than
for which
is an integer. Find
.
Solution
The best way to solve this problem is to get the iterated part out of the exponent:
Since
, we can easily use induction to show that
. So now we only need to find the next value of
that makes
an integer. This means that
must be a power of
. We test
:
This has no integral solutions, so we try
:
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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All AIME Problems and Solutions |