2013 AMC 8 Problems/Problem 10: Difference between revisions
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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> | 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 | == Video Solution by WhyMath == | ||
https://youtu.be/CWZkTCNu42o?t=31 | https://youtu.be/CWZkTCNu42o?t=31 | ||
~ | ~ Savannahsolver | ||
==Video Solution== | ==Video Solution by OmegaLearn== | ||
https://youtu.be/ZtjQTa2hWmw ~ | https://youtu.be/ZtjQTa2hWmw ~ pi is 3.14 | ||
==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 23:28, 3 November 2025
Problem
What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
Solution 1
To find either the LCM or the GCF of two numbers, always prime factorize first.
The prime factorization of
.
The prime factorization of
.
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
). 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.
= 18.
Thus the answer =
=
.
Solution 2
We start off with a similar approach as the original solution. From the prime factorizations, the GCF is
.
It is a well known fact that
. So we have,
.
Dividing by
yields
.
Therefore,
.
Solution 3
From Solution 1,
the prime factorization of
.
The prime factorization of
.
Hence,
, and
.
Therefore,
Video Solution by WhyMath
https://youtu.be/CWZkTCNu42o?t=31
~ Savannahsolver
Video Solution by OmegaLearn
https://youtu.be/ZtjQTa2hWmw ~ pi is 3.14
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing