Question

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

$32$

$36$

$37$

$39$

$41$

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 $(0$ through $8,)$ and $4$ options for multiples of $3$ $(0$ through $9).$

This gives us $5^2$ options for the squares and $4^2$ options for the cubes. We have to subtract out the powers of $6,$ 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.