Difference between revisions of "2013 AMC 8 Problems/Problem 10"
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The prime factorization of <math>594 = 3^3 \times 11 \times 2</math>. | The prime factorization of <math>594 = 3^3 \times 11 \times 2</math>. | ||
− | 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. | + | 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. | 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. |
Revision as of 08:33, 19 February 2016
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 = = .
Similar Solution
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, .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.