2012 AMC 10B Problems/Problem 10

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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

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 10 Problems and Solutions

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