# 2012 AMC 10B Problems/Problem 10

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## Problem

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 1

Cross-multiplying gives $MN=36.$ We write $36$ as a product of two positive integers: \begin{align*} 36 &= 1\cdot36 \\ &= 2\cdot18 \\ &= 3\cdot12 \\ &= 4\cdot9 \\ &= 6\cdot6. \end{align*} The products $1\cdot36, 2\cdot18, 3\cdot12,$ and $4\cdot9$ each produce $2$ ordered pairs $(M,N),$ as we can switch the order of the factors. The product $6\cdot6$ produces $1$ ordered pair $(M,N).$ Together, we have $4\cdot2+1=\boxed{\textbf{(D)}\ 9}$ ordered pairs $(M,N).$

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## Solution 2

Cross-multiplying gives $MN=36.$ From the prime factorization $$36=2^2\cdot3^2,$$ we conclude that $36$ has $(2+1)(2+1)=9$ positive divisors. There are $9$ values of $M,$ and each value generates $1$ ordered pair $(M,N).$ So, there are $\boxed{\textbf{(D)}\ 9}$ ordered pairs $(M,N)$ in total.

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