2012 AMC 10B Problems/Problem 10

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Problem 10

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

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\10$ (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*36=36

2*18=36

3*12=36

4*9=36

6*6=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.

$4*2+1=\boxed{9}$

OR

$\textbf{(D)}$

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