2019 AMC 10A Problem 11

Question

How many positive integer divisors of

201^9

are perfect squares or perfect cubes (or both)?

32

36

37

39

41

Factor

201

first

Use inclusion-exclusion on divisor exponents that are even or multiples of

3

Taking the prime factorization of

201^9,

we get

3^9 \cdot 67^9.

Note that a perfect square has even exponents for its prime factors, and a cube’s exponents are divisible by

3.

There are

5

options for an even exponent, from

0

through

8,

and

4

options for multiples of

3,

from

0

through

9.

This gives us

5^2

options for the squares and

4^2

options for the cubes. We have to subtract out the sixth powers, however.

Using the same logic, sixth powers have to have exponents of prime factors be divisible by

6.

There are

2

options,

0

and

6.

This means that there are

2^2 = 4

sixth powers. This gives us a total of

25 + 16 - 4 = 37.

perfect squares or perfect cubes.

Thus, C is the correct answer.

Problem 11 in other years