2019 AMC 10A Problems/Problem 11

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

Solution by Aadileo

Solution 2

201 = 67 * 3. Split the answers into cases:

Case 1: The factor is 3^n

Then, we would have n = 2, 3, 4, 6, 8, and 9.

Case 2: The factor is 67^n

Same as case 1

Case 3: The factor is some combination

This would be easy if we could just have any combination, as that would be simply 6*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 n values.

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

n = 3 is a "cube" because it would be 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 its both a "square" and a "cube". Don't count this as either because it's double function would double count the numbers if we counted them simply as a "square" or as a "cube", so we will count them in another case.

n = 8 is a "square"

n = 9 is a "cube".

Now let's do 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*3 = 9 cases

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*2 = 4 cases

Subcase 3: A number pairs with n=6.

Since any number can pair with n=6 since it's a square AND cube, there are 6 cases. Remember however that there can be two different bases (3 and 67), and they would produce different results. Thus, there are 6*2 = 12 cases

Thus, we add up the cases, to get 6+6+9+4+12 = 37.

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2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2019 AMC 10A Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

See Also