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

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\textbf{(D)}\    165  \qquad
 
\textbf{(D)}\    165  \qquad
 
\textbf{(E)}\    610</math>
 
\textbf{(E)}\    610</math>
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==Solution==
 +
Take the prime factorization of <math>360</math>. <math>360=2^3*3^2*5</math>.
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We want <math>x</math> to be as small as possible. And you want <math>x*360</math> to be a square.
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So <math>x=2*5=10</math>.
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<math>y</math> is simmilar. <math>y=3*5^2=3*25=75</math>
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So, <math>x+y=75+10=85</math>, or <math>\boxed{\textbf{(B)}\ 85}</math>.
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 +
~ModestFox97
  
 
==Video Solution (XXX)==
 
==Video Solution (XXX)==
 
https://www.youtube.com/watch?v=ZuSJdf1zWYw  ~David
 
https://www.youtube.com/watch?v=ZuSJdf1zWYw  ~David
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−
==Solution==
 
−
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>.
 
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==:)==
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=16|num-a=18}}
 
{{AMC8 box|year=2010|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:26, 14 August 2025

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 (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

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

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