Difference between revisions of "2019 AMC 10A Problems/Problem 11"
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How many positive integer divisors of <math>201^9</math> are perfect squares or perfect cubes (or both)? | How many positive integer divisors of <math>201^9</math> are perfect squares or perfect cubes (or both)? | ||
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Solution by Aadileo | Solution by Aadileo | ||
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| + | ==See Also== | ||
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| + | {{AMC10 box|year=2019|ab=A|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:36, 9 February 2019
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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