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

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==Solution==
 
==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 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 <math>y</math> is <math>3*5^2=75</math>. Thus, our answer is <math>x+y=10+75=\boxed{\textbf{(B)}\ 85}</math>.
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The prime factorization of <math>360=2^3 \cdot 3^2 \cdot 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 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 <math>y</math> is <math>3 \cdot 5^2=75</math>. Thus, our answer is <math>x+y=10+75=\boxed{\textbf{(B)}\ 85}</math>.
  
 
==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}}
 
{{MAA Notice}}

Revision as of 17:20, 15 September 2018

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

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
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

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