Question

$201^9$ 的正整数因子中，有多少个是完全平方数或完全立方数（或两者都是）？

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

$32$

$36$

$37$

$39$

$41$

对 $201^9$ 作质因数分解，得到 $3^9 \cdot 67^9$ 。

注意完全平方数的质因数指数为偶数，完全立方数的质因数指数能被 $3$ 整除。

偶数指数有 $5$ 种选择，从 $0$ 到 $8$ ； $3$ 的倍数指数有 $4$ 种选择，从 $0$ 到 $9$ 。

因此平方数有 $5^2$ 种选择，立方数有 $4^2$ 种选择；但还要减去重复计数的六次幂。

同理，六次幂的质因数指数必须能被 $6$ 整除。每个指数有 $2$ 种选择，即 $0$ 和 $6$ 。

六次幂共有 $2^2 = 4$ 个，所以由容斥原理，所求数量为 $25 + 16 - 4 = 37.$

所以正确答案是 C。

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.