Difference between revisions of "2012 AMC 10B Problems/Problem 10"
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Now you can reverse the order of the factors for all of the ones listed above, because they are ordered pairs except for 6*6 since it is the same back if you reverse the order. | Now you can reverse the order of the factors for all of the ones listed above, because they are ordered pairs except for 6*6 since it is the same back if you reverse the order. | ||
| − | <math>4*2+1= | + | <math>4*2+1=9</math> |
| − | + | \boxed{\textbf{(B)}\ 9}$ | |
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Revision as of 20:31, 17 February 2013
Problem 10
How many ordered pairs of positive integers (M,N) satisfy the equation 
= 
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\10$ (Error compiling LaTeX. Unknown error_msg)
Solution
=
is a ratio; therefore, you can cross-multiply.
Now you find all the factors of 36:
.
Now you can reverse the order of the factors for all of the ones listed above, because they are ordered pairs except for 6*6 since it is the same back if you reverse the order.
\boxed{\textbf{(B)}\ 9}$
