2019 AMC 10A Problems/Problem 11
Contents
Problem
How many positive integer divisors of are perfect squares or perfect cubes (or both)?
Solution 1 (PIE)
Prime factorizing , we get . A perfect square must have even powers of its prime factors, so our possible choices for our exponents to get 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 ( , , , and ), we get .
Solution 2
Observe that . Now divide into cases:
Case 1: The factor is . Then we can have , , , , , or .
Case 2: The factor is . This is the same as Case 1.
Case 3: The factor is some combination of s and s.
This would be easy if we could just have any combination, as that would simply give . 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 .
is a "square" because it would give a factor of this number that is a perfect square. More generally, it is even.
is a "cube" because it would give a factor of this number that is a perfect cube. More generally, it is a multiple of .
is a "square".
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.
is a "square"
is a "cube".
Now let's consider subcases:
Subcase 1: The squares are with each other.
Since we have square terms, and they would pair with other square terms, we get possibilities.
Subcase 2: The cubes are with each other.
Since we have cube terms, and they would pair with other cube terms, we get possibilities.
Subcase 3: A number pairs with .
Since any number can pair with (as it gives both a square and a cube), there would be possibilities. Remember however that there can be two different bases ( and ), and they would produce different results. Thus, there are in fact possibilities.
Finally, summing the cases gives .
Solution 3 (Quick)
We first prime factorize . Then, to get a perfect square, we must have an even number in the exponent. To get an odd cube, we must have a multiple of in the exponent. The largest square for can be , so their must be ways. The largest cube is , so there must be . Minus one due to overlapping and we get ways for to be a cube/square. We can see that this same thing happens for due to the same exponent. Adding as a case, we have our answer;
~ Wiselion
Solution 4 (A Little Long)
Notice that . We factorize to get . We then list perfect squares and cubes. , , , . , , . , , , . , , . Notice that the powers of overlap. We must not forget though. Of course, all of these factors already work. This gives us . Next, we count the perfect squares. Since there are options we have . We do the same for the perfect cubes except with 3 options this time, and we have . However, we accidentally overcounted . We add our answers and subtract to get
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Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/ZhAZ1oPe5Ds?t=2402
~ pi_is_3.14
Video Solution
Education, the Study of Everything
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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