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

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\textbf{(E)}\    610</math>
 
\textbf{(E)}\    610</math>
  
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==Solution==
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The prime factorization of <math>360=2^3*3^2*5</math>. 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 <math>y=3*5^2=75</math>. Thus x+y=85, B.
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=16|num-a=18}}
 
{{AMC8 box|year=2009|num-b=16|num-a=18}}

Revision as of 14:55, 11 December 2012

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=85, B.

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AJHSME/AMC 8 Problems and Solutions
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