# 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. ! Undefined control sequence.)

## 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}$