2016 AIME II Problems/Problem 11
Contents
Problem
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 and firebolt360
Solution 2
All integers will have factorization . Therefore, the number of factors in is , and for is . The most salient step afterwards is to realize that all numbers not and also not satisfy the criterion. The cycle repeats every integers, and by PIE, of them are either -nice or -nice or both. Therefore, we can take numbers minus the that work between inclusive, to get positive integers less than that are not nice for .
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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