Difference between revisions of "2016 AIME II Problems/Problem 11"
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Revision as of 22:34, 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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