2009 AIME I Problems/Problem 7
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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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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