2019 AMC 10A Problems/Problem 11
Contents
Problem
How many positive integer divisors of 
are perfect squares or perfect cubes (or both)?
Solution 1
Prime factorizing , we get 
.
A perfect square must have even powers of its prime factors, so our possible choices for our exponents of a perfect square are 
for both 
and 
. This yields 
perfect squares.
Perfect cubes must have multiples of for each of their prime factors' exponents, so we have either 
, or 
for both and , which yields 
perfect cubes, for a total of 
.
Subtracting the overcounted powers of six (
, 
, 
, and 
), we get 
.
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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See Also
| 2019 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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