2013 UNCO Math Contest II Problems/Problem 2

Problem

A number $x$ is equal to $7\cdot24\cdot48$ . What is the smallest positive integer $y$ such that the product $xy$ is a perfect cube?

Solution

We can factor $12$ into $2 \times 2 \times 3$ . 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 $2 \times 12$ , which is $24$ . Because we only have one power of three, we need two more, so we multiply $24 \times 3 \times 3$ , giving us $216$ , which is a perfect cube.

To find a perfect 6th power, we multiply $36$ by $216$ to get $7776$ . We know that the factorization of this number is $2^5 \times 3^5$ . This means that this number is $6^5$ . We need a perfect 6th, so we multiply by $6$ to get $46656$ , which is $6^6$ , or $\boxed{588}$ .

2013 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2013 UNCO Math Contest II Problems/Problem 2

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