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 $y$ ?

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

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=2768

Video Solution

https://www.youtube.com/watch?v=ZuSJdf1zWYw

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 $x+y=10+75=\boxed{\textbf{(B)}\ 85}$ .

Solution 2 (Using Answer Choices)

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 $75+10=85$ .

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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