Difference between revisions of "2016 AIME II Problems/Problem 11"
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Revision as of 21:34, 16 May 2016
For positive integers and , define to be -nice if there exists a positive integer such that has exactly positive divisors. Find the number of positive integers less than that are neither -nice nor -nice.
Solution
We claim that an integer is only -nice if and only if . By the number of divisors formula, the number of divisors of is . Since all the s are divisible by in a perfect power, the only if part of the claim follows. To show that all numbers are -nice, write . Note that has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than that are either or is , so the desired answer is .
Solution by Shaddoll
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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