2013 UNCO Math Contest II Problems/Problem 2

Problem

EXAMPLE: The number is equal to and also equal to , so is both a perfect square and a perfect cube.

(a) Find the smallest positive integer multiple of that is a perfect square.

(b) Find the smallest positive integer multiple of that is a perfect cube.

(c) Find the smallest positive integer multiple of that is both a perfect square and a perfect cube.

Solution

We can factor 12 into 2*2*3. There are already two factors of two, so we only need to multiply it by 3 to get two factors of three, giving us 36.

To find the perfect cube, we need all of the prime factors to be to the third power. Because 2 is squared, we need to multiply by a power of 2, giving us 2*12=24. Because we only have one power of three, we need two more, so we multiply 24*3*3, giving us 216, which is a perfect cube.

To find a perfect 6th power, we multiply 36*216 to get 7776. We know that the factorization of this number is 2^5*3^5. This means that this number is 6^5. We need a perfect 6th, so we multiply by 6 to get 46656, which is 6^6.