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

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From the question's requirements, we can figure out <math>x</math> is 10. Then we can use the answer choices to find what <math>y</math> is.  
 
From the question's requirements, we can figure out <math>x</math> is 10. Then we can use the answer choices to find what <math>y</math> is.  
 
Let's start with <math>A</math>. If <math>A</math> was right, then <math>y=70</math>. We can multiply <math>70</math> by <math>360</math> and get <math>25200</math>, which isn't a perfect cube. Then we move to <math>B</math>. <math>85-10=75</math>, so <math>y=75</math> if <math>B</math> is right. Then we multiply <math>75</math> by <math>360</math> to get <math>27000</math>, which is <math>30^3</math>. Therefore, our answer is <math>\boxed{\textbf{(B)}\ 85}</math> because <math>y=75</math> and <math>75+10=85</math>.
 
Let's start with <math>A</math>. If <math>A</math> was right, then <math>y=70</math>. We can multiply <math>70</math> by <math>360</math> and get <math>25200</math>, which isn't a perfect cube. Then we move to <math>B</math>. <math>85-10=75</math>, so <math>y=75</math> if <math>B</math> is right. Then we multiply <math>75</math> by <math>360</math> to get <math>27000</math>, which is <math>30^3</math>. Therefore, our answer is <math>\boxed{\textbf{(B)}\ 85}</math> because <math>y=75</math> and <math>75+10=85</math>.
 +
 +
~Trex226
  
 
==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 14:04, 10 January 2023

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 $y$?

$\textbf{(A)}\ 80 \qquad \textbf{(B)}\ 85 \qquad \textbf{(C)}\ 115 \qquad \textbf{(D)}\ 165 \qquad \textbf{(E)}\ 610$

Video Solution by OmegaLearn

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

Video Solution

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

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

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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