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

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(Solution)
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<math>\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660</math>
 
<math>\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660</math>
  
==Solution==
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==Solution 1==
 
This is very easy. To find the LCM of 180 and 594, first find the prime factorization of both.
 
This is very easy. To find the LCM of 180 and 594, first find the prime factorization of both.
  
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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>
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==Similar Solution==
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We start off with a similar approach as the original solution. From the prime factorizations, the GCF is <math>18</math>.
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It is a well known fact that <math>\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|</math>. So we have, <math>18\times \operatorname{lcm} (180,594)=594\times 180</math>.
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Dividing by <math>18</math> yields <math>\operatorname{lcm} (180,594)=594\times 10=5940</math>.
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Therefore, <math>\dfrac{\operatorname{lcm} (180,594)}{\gcd(180,594)}=\dfrac{5940}{18}=\boxed{\textbf{(C)}\ 330}</math>.
  
 
==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 19:44, 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 1

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

Similar Solution

We start off with a similar approach as the original solution. From the prime factorizations, the GCF is $18$.

It is a well known fact that $\gcd(m,n)\times \operatorname{lcm}(m,n)=|mn|$. So we have, $18\times \operatorname{lcm} (180,594)=594\times 180$.

Dividing by $18$ yields $\operatorname{lcm} (180,594)=594\times 10=5940$.


Therefore, $\dfrac{\operatorname{lcm} (180,594)}{\gcd(180,594)}=\dfrac{5940}{18}=\boxed{\textbf{(C)}\ 330}$.

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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