2012 AMC 10B Problems/Problem 10

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 $(M,N).$

~Rguan (Solution)

~MRENTHUSIASM (Reformatting)

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 in total.

~MRENTHUSIASM

2012 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2012 AMC 10B Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

See Also