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2019 AMC 10A Problem 11

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2019 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 3 of 5 · about 3 min · position P11-15

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

A)
$32$
B)
$36$
C)
$37$
D)
$39$
E)
$41$

Answer

Key insight

Since 201 = 3 x 67, divisors are 3^a 67^b; squares need both exponents even (25), cubes need multiples of 3 (16), overlap is sixth powers (4).

Solution

Since $201 = 3 \cdot 67$ , we have $201^9 = 3^9 \cdot 67^9$ , and every divisor has the form $3^a 67^b$ with $0 \le a, b \le 9$ .

Perfect squares: $a$ and $b$ both even. The even values in $0, \dots, 9$ are $0, 2, 4, 6, 8$ , five choices each, giving $5 \cdot 5 = 25$ .
Perfect cubes: $a$ and $b$ both multiples of $3$ : $0, 3, 6, 9$ , four choices each, giving $4 \cdot 4 = 16$ .
Both (perfect sixth powers): $a, b \in \{0, 6\}$ , giving $2 \cdot 2 = 4$ .

By inclusion-exclusion, the count is $25 + 16 - 4 = 37$ .

The answer is $\boxed{\textbf{(C)}\ 37}$ .

Why this works

A divisor of $p^m q^n$ is a perfect $k$ th power exactly when each exponent is a multiple of $k$ , so counting such divisors is counting exponent pairs, and "square or cube" is a union of two sets whose intersection is the sixth powers. Always factor the base first: $201$ is not prime, and its two prime factors are what make the count two-dimensional.

The trap

Forgetting to subtract the divisors that are both squares and cubes (sixth powers), which gives 41, or treating 201 as prime.

Common mistakes

Forgetting to subtract the divisors that are both squares and cubes (sixth powers), which gives 41, or treating 201 as prime.
Counting even exponents as $4$ (omitting $0$ ) or multiples of $3$ as $3$ (omitting $9$ or $0$ ), which produces $32$ or $36$ .

Techniques

Set up the equation/formula and compute; no special trick needed