2019 AMC 10A Problems/Problem 11
Problem
How many positive integer divisors of are perfect squares or perfect cubes (or both)?
Solution 1
Prime factorizing , we get
.
A perfect square must have even powers of its prime factors, so our possible choices for our exponents of a perfect square are
for both
and
. This yields
perfect squares.
Perfect cubes must have multiples of for each of their prime factors' exponents, so we have either
, or
for both
and
, which yields
perfect cubes, for a total of
.
Subtracting the overcounted powers of six ( ,
,
, and
), we get
.
Solution by Aadileo
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AMC 10 Problems and Solutions |
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