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

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Thus the answer = <math>\frac{5940}{18}</math> = <math>\boxed{\textbf{(C)}\ 330}</math>
 
Thus the answer = <math>\frac{5940}{18}</math> = <math>\boxed{\textbf{(C)}\ 330}</math>
 
--[[User:Arpanliku|Arpanliku]] 16:48, 27 November 2013 (EST) Sorry, I'm an amateur at LATEX.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=9|num-a=11}}
 
{{AMC8 box|year=2013|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:35, 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 = $\frac{5940}{18}$ = $\boxed{\textbf{(C)}\ 330}$

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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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