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Difference between revisions of "2013 AMC 8 Problems/Problem 10"

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The prime factorization of 180 = <math>3^2 \times  5 \times 2^2</math>
 
The prime factorization of 180 = <math>3^2 \times  5 \times 2^2</math>
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The prime factorization of 594 = <math>3^3 \times  11 \times 2</math>
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The prime factorization of 594 = <math>3^3 \times  11 \times 2</math> Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is <math>3^3, 5, 11, 2^2</math>). Multiply all of these to get 5940.  
−
 
 
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Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is <math>3^3, 5, 11, 2^2</math>). Multiply all of these to get 5940.  
 
  
 
For the GCF of 180 and 594, use the least power of all of the numbers THAT ARE IN BOTH and multiply. <math>3^2 \times 2</math> = 18.
 
For the GCF of 180 and 594, use the least power of all of the numbers THAT ARE IN BOTH and multiply. <math>3^2 \times 2</math> = 18.

Revision as of 16:49, 27 November 2013

Problem

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?

$\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660$

Solution

This is very easy. To find the LCM of 180 and 594, first find the prime factorization of both.

The prime factorization of 180 = $3^2 \times 5 \times 2^2$ The prime factorization of 594 = $3^3 \times 11 \times 2$ Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is $3^3, 5, 11, 2^2$). Multiply all of these to get 5940.

For the GCF of 180 and 594, use the least power of all of the numbers THAT ARE IN BOTH and multiply. $3^2 \times 2$ = 18.

Thus the answer = $\frac{5940}{18}$ = $\boxed{\textbf{(C)}\ 330}$

--Arpanliku 16:48, 27 November 2013 (EST) Sorry, I'm an amateur at LATEX.

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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