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

m (Solution)
(Solution)
Line 11: Line 11:
 
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, 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 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 and multiply. <math>3^2 \times 2</math> = 18.

Revision as of 19:36, 27 November 2013

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

This is very easy. To find the LCM of 180 and 594, first find the prime factorization of both.

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

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

Then, 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 and multiply. $3^2 \times 2$ = 18.

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS