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

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

@@ Line 1: / Line 1: @@
 ==Problem==
+What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
-==Solution==
+<math>\textbf{(A)}\ 110 \qquad \textbf{(B)}\ 165 \qquad \textbf{(C)}\ 330 \qquad \textbf{(D)}\ 625 \qquad \textbf{(E)}\ 660</math>
+==Solution 1==
+To find either the LCM or the GCF of two numbers, always prime factorize first.
+The prime factorization of <math>180 = 3^2 \times  5 \times 2^2</math>.
+The prime factorization of <math>594 = 3^3 \times  11 \times 2</math>.
+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.
+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.
+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==
-{{AMC8 box|year=2013|before=First Problem|num-a=11}}
+{{AMC8 box|year=2013|num-b=9|num-a=11}}
 {{MAA Notice}}

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