2013 AMC 8 Problems/Problem 10

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, $\frac{\operatorname{lcm} (180,594)}{\operatorname{gcf} (180,594)} = \frac{2^2 \cdot 3^3 \cdot 5 \cdot 11}{2 \cdot 3^2} = 2 \cdot 3 \cdot 5 \cdot 11 = 330 \Longrightarrow \boxed{\textbf{(C)}\ 330}$

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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