2019 AMC 10A Problems/Problem 11

Problem

How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)?

${\textbf{(A) }32} \qquad {\textbf{(B) }36} \qquad {\textbf{(C) }37} \qquad {\textbf{(D) }39} \qquad {\textbf{(E) }41}$

Solution 1

Prime factorizing $201^9$ , we get $3^9\cdot67^9$ . A perfect square must have even powers from its prime factors, so our possible choices for our exponents of a perfect square are $0, 2, 4, 6, 8$ for both $3$ and $67$ . This yields $5\cdot5 = 25$ perfect squares. Perfect cubes must have multiples of 3 for each of their prime factors' exponents, so we have either $0, 3, 6$ , or $9$ for both $3$ and $67$ , which yields $4\cdot4 = 16$ perfect cubes, for a total of $25+16 = 41$ . Subtracting the overcounted powers of six ( $3^0\cdot67^0$ , $3^0\cdot67^6$ , $3^6\cdot67^0$ , and $3^6\cdot67^6$ ), we get $41-4 = \boxed{\textbf{(C) }37}$ .

Solution by Aadileo

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2019 AMC 10A Problems/Problem 11

Problem

Solution 1

See Also