Difference between revisions of "2019 AMC 10A Problems/Problem 11"
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Finally, summing the cases gives <math>6+6+9+4+12 = \boxed{\textbf{(C) }37}</math>. | Finally, summing the cases gives <math>6+6+9+4+12 = \boxed{\textbf{(C) }37}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/JR1LpMc3Ntg | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 20:43, 19 October 2020
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 ( , , , 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 .
Video Solution
~savannahsolver
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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