2012 AMC 10B Problems/Problem 10

Revision as of 20:31, 17 February 2013 by RainbowsHurt (talk | contribs) (Solution)

Problem 10

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

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\10$ (Error compiling LaTeX. ! Undefined control sequence.)

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.

$4*2+1=9$

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

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