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 from 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 3 for each of their prime factors' exponents, so we have either
, or
for both
and
, which yields
perfect cubes. In total, we have
. However, we have overcounted perfect 6'ths:
,
,
, and
. We must subtract these 4, for our final answer, which is
.
Solution by Aadileo
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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