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

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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>
  
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==Solution==
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==Solution 1==  
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This is very easy. To find the LCM of 180 and 594, first find the prime factorization of both.
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To find either the LCM or the GCF of two numbers, always prime factorize first.
  
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The prime factorization of 180 = <math>3^2 \times  5 \times 2^2</math>
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The prime factorization of <math>180 = 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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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.  
+
The prime factorization of <math>594 = 3^3 \times  11 \times 2</math>.
  
−
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.
+
Then, to find the LCM, we have to 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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Thus the answer = <math>\frac{5940}{18}</math> = <math>\boxed{\textbf{(C)}\ 330}</math>
+
For the GCF of 180 and 594, use the least power of all of the numbers that are in both factorizations and multiply. <math>3^2 \times 2</math> = 18.
  
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--[[User:Arpanliku|Arpanliku]] 16:48, 27 November 2013 (EST) Sorry, I'm an amateur at LATEX.
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Thus the answer = <math>\frac{5940}{18}</math> = <math>\boxed{\textbf{(C)}\ 330}</math>.
 +
 
 +
==Solution 2==
 +
We start off with a similar approach as the original solution. From the prime factorizations, the GCF is <math>18</math>.
 +
 
 +
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>.
 +
 
 +
Dividing by <math>18</math> yields <math>\operatorname{lcm} (180,594)=594\times 10=5940</math>.
 +
 
 +
Therefore, <math>\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf}(180,594)}=\frac{5940}{18}=\boxed{\textbf{(C)}\ 330}</math>.
 +
 
 +
==Solution 3==
 +
From Solution 1,
 +
the prime factorization of <math>180 = 2^2 \cdot 3^2 \cdot 5</math>.
 +
The prime factorization of <math>594 = 2 \cdot 3^3 \cdot 11</math>.
 +
Hence, <math>\operatorname{lcm} (180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11</math>, and <math>\operatorname{gcf} (180,594) = 2 \cdot 3^2</math>.
 +
Therefore, <math>\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 = 330 \Longrightarrow \boxed{\textbf{(C)}\ 330}</math>
 +
 
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/CWZkTCNu42o?t=31
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution==
 +
https://youtu.be/ZtjQTa2hWmw ~savannahsolver
  
 
==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}}

Latest revision as of 21:42, 5 January 2024

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

To find either the LCM or the GCF of two numbers, always prime factorize first.

The prime factorization of $180 = 3^2 \times 5 \times 2^2$.

The prime factorization of $594 = 3^3 \times 11 \times 2$.

Then, to find the LCM, we have to 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 factorizations and multiply. $3^2 \times 2$ = 18.

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

Solution 2

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, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf}(180,594)}=\frac{5940}{18}=\boxed{\textbf{(C)}\ 330}$.

Solution 3

From Solution 1, the prime factorization of $180 = 2^2 \cdot 3^2 \cdot 5$. The prime factorization of $594 = 2 \cdot 3^3 \cdot 11$. Hence, $\operatorname{lcm} (180,594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11$, and $\operatorname{gcf} (180,594) = 2 \cdot 3^2$. Therefore, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 = 330 \Longrightarrow \boxed{\textbf{(C)}\ 330}$

Video Solution by OmegaLearn

https://youtu.be/CWZkTCNu42o?t=31

~ pi_is_3.14

Video Solution

https://youtu.be/ZtjQTa2hWmw ~savannahsolver

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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