2019 AMC 10A Problems/Problem 11

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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 of 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 $6$ ( $3^0\cdot67^0$ , $3^0\cdot67^6$ , $3^6\cdot67^0$ , and $3^6\cdot67^6$ ), we get $41-4 = \boxed{\textbf{(C) }37}$ .

Solution 2

Observe that $201 = 67 \cdot 3$ . Now divide into cases:

Case 1: The factor is $3^n$ . Then we can have $n = 2$ , $3$ , $4$ , $6$ , $8$ , or $9$ .

Case 2: The factor is $67^n$ . This is the same as Case 1.

Case 3: The factor is some combination of $3$ s and $67$ s.

This would be easy if we could just have any combination, as that would simply give $6 \cdot 6$ . However, we must pair the numbers that generate squares with the numbers that generate squares and the same for cubes. In simpler terms, let's organize our values for $n$ .

$n = 2$ is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.

$n = 3$ is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of $3$ .

$n = 4$ is a "square".

$n = 6$ is interesting, since it's both a "square" and a "cube". Don't count this as either because this would double-count, so we will count this in another case.

$n = 8$ is a "square"

$n = 9$ is a "cube".

Now let's consider subcases:

Subcase 1: The squares are with each other.

Since we have $3$ square terms, and they would pair with $3$ other square terms, we get $3 \cdot 3 = 9$ possibilities.

Subcase 2: The cubes are with each other.

Since we have $2$ cube terms, and they would pair with $2$ other cube terms, we get $2 \cdot 2 = 4$ possibilities.

Subcase 3: A number pairs with $n=6$ .

Since any number can pair with $n=6$ (as it gives both a square and a cube), there would be $6$ possibilities. Remember however that there can be two different bases ( $3$ and $67$ ), and they would produce different results. Thus, there are in fact $6 \cdot 2 = 12$ possibilities.

Finally, summing the cases gives $6+6+9+4+12 = \boxed{\textbf{(C) }37}$ .

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2019 AMC 10A (Problems • Answer Key • Resources)
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2019 AMC 10A Problems/Problem 11

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