2009 AMC 8 Problems/Problem 17

Problem

The positive integers $x$ and $y$ are the two smallest positive integers for which the product of $360$ and $x$ is a square and the product of $360$ and $y$ is a cube. What is the sum of $x$ and $y$?

$\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 85 \qquad \textbf{(C)}\ 115 \qquad \textbf{(D)}\ 165 \qquad \textbf{(E)}\ 610$

Solution

The prime factorization of $360=2^3 \cdot 3^2 \cdot 5$. If a number is a perfect square, all of the exponents in its prime factorization must be even. Thus we need to multiply by a 2 and a 5, for a product of 10, which is the minimum possible value of x. Similarly, y can be found by making all the exponents divisible by 3, so the minimum possible value of $y$ is $3 \cdot 5^2=75$. Thus, our answer is $x+y=10+75=\boxed{\textbf{(B)}\ 85}$.

From the question's requirements, we can figure out $x$ is $10$. Then we can use the answer choices to find what $y$ is. Let's start with $A$. If $A$ was right, then $y=70$. We can multiply $70$ by $360$ and get $25200$, which isn't a perfect cube. Then we move to $B$. $85-10=75$, so $y=75$ if $B$ is right. Then we multiply $75$ by $360$ to get $27000$, which is $30^3$. Therefore, our answer is $\boxed{\textbf{(B)}\ 85}$ because $y=75$ and $75+10=85$.