2009 AMC 8 Problems/Problem 17

Revision as of 18:26, 14 August 2025 by Modestfox97 (talk | contribs) (Video Solution (XXX))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

Take the prime factorization of $360$. $360=2^3*3^2*5$. We want $x$ to be as small as possible. And you want $x*360$ to be a square. So $x=2*5=10$. $y$ is simmilar. $y=3*5^2=3*25=75$ So, $x+y=75+10=85$, or $\boxed{\textbf{(B)}\ 85}$.

~ModestFox97

Video Solution (XXX)

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

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png