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# Difference between revisions of "2012 AMC 10B Problems/Problem 10"

## Problem 10

How many ordered pairs of positive integers (M,N) satisfy the equation $\frac {M}{6}$ = $\frac{6}{N}$

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

## Solution

$\frac {M}{6}$ = $\frac{6}{N}$

is a ratio; therefore, you can cross-multiply.

$MN=36$

Now you find all the factors of 36:

$1\times36=36$

$2\times18=36$

$3\times12=36$

$4\times9=36$

$6\times6=36$.

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.

$4\cdot 2+1=9$

$\boxed{\textbf{(D)}\ 9}$

## See Also

 2012 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 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 AMC 10 Problems and Solutions

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

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