2012 AMC 10B Problem 10

1. Cross-multiply $\frac{M}{6} = \frac{6}{N}$ to get $MN = 36$.
2. So we need two positive whole numbers that multiply to $36$. Once $M$ is picked, $N = 36 \div M$ is fixed, and it is a whole number only when $M$ divides $36$.
3. List the divisors of $36$ in factor pairs: $1 \times 36$, $2 \times 18$, $3 \times 12$, $4 \times 9$, $6 \times 6$.
4. The pairs are ordered, so $(4, 9)$ and $(9, 4)$ count separately. Each of the first four factor pairs gives two ordered pairs, and $6 \times 6$ gives only one:
$$(1,36), (36,1), (2,18), (18,2), (3,12), (12,3), (4,9), (9,4), (6,6).$$
5. That is $4 \times 2 + 1 = 9$ ordered pairs.

The answer is $\boxed{\textbf{(D)}\ 9}$.

An equation like $\frac{M}{6} = \frac{6}{N}$ is really a product equation, $MN = 36$. The number of ordered solutions equals the number of divisors of $36$. A quick way to count divisors: $36 = 2^2 \times 3^2$, and a divisor can use $0$, $1$ or $2$ twos and $0$, $1$ or $2$ threes, so there are $3 \times 3 = 9$ divisors.

Cross-multiplying turns the equation into $MN = 36$. Choosing any positive divisor $M$ of $36$ determines $N = 36/M$, so the ordered pairs correspond exactly to the positive divisors of $36$.

Since $36 = 2^2 \cdot 3^2$, the number of divisors is $(2+1)(2+1) = 9$. Listing them: $M \in \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$, giving the pairs $(1,36), (2,18), (3,12), (4,9), (6,6), (9,4), (12,3), (18,2), (36,1)$.

The answer is $\boxed{\textbf{(D)}\ 9}$.