Difference between revisions of "2009 AMC 8 Problems/Problem 17"

(Solution)
Line 14: Line 14:
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=16|num-a=18}}
 
{{AMC8 box|year=2009|num-b=16|num-a=18}}
 +
{{MAA Notice}}

Revision as of 00:49, 5 July 2013

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 $360=2^3*3^2*5$. 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 x. Similarly, y can be found by making all the exponents divisible by 3, so $y=3*5^2=75$. Thus $x+y=\boxed{\textbf{(B)}\ 85}$.

See Also

2009 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

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