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

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The positive integers <math> x</math> and <math> y</math> are the two smallest positive integers for which the product of <math> 360</math> and <math> x</math> is a square and the product of <math> 360</math> and <math> y</math> is a cube. What is the sum of <math> x</math> and <math> y</math>?
 
The positive integers <math> x</math> and <math> y</math> are the two smallest positive integers for which the product of <math> 360</math> and <math> x</math> is a square and the product of <math> 360</math> and <math> y</math> is a cube. What is the sum of <math> x</math> and <math> y</math>?
  
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<math> \textbf{(A)}\  80   \qquad
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<math> \textbf{(A)}\  8   \qquad
 
\textbf{(B)}\    85  \qquad
 
\textbf{(B)}\    85  \qquad
 
\textbf{(C)}\    115  \qquad
 
\textbf{(C)}\    115  \qquad

Revision as of 22:00, 17 January 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 $y$?

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 85 \qquad \textbf{(C)}\ 115 \qquad \textbf{(D)}\ 165 \qquad \textbf{(E)}\ 610$

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=2768

Video Solution (XXX)

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

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

Solution 2 (Using Answer Choices)

From the question's requirements, we can figure out $x$ is $10$. Then we can use the answer choices to find what $y$ is. Let's start with $A$. If $A$ was right, then $y=70$. We can multiply $70$ by $360$ and get $25200$, which isn't a perfect cube. Then we move to $B$. $85-10=75$, so $y=75$ if $B$ is right. Then we multiply $75$ by $360$ to get $27000$, which is $30^3$. Therefore, our answer is $\boxed{\textbf{(B)}\ 85}$ because $y=75$ and $75+10=85$.

~Trex226

:)

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

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