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: Plug in to see the first few terms of the sequence: We notice that the terms are in arithmetic progression. 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 :
Solution 2 (Telescoping)
We notice that by multiplying the equation from an arbitrary all the way to , we get: This simplifies to We can now test powers of .
- that gives us , which is useless.
- that gives a non-integer .
- that gives .
-integralarefun
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 | ||
All AIME Problems and Solutions |
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