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2012 AMC 10B Problem 10

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2012 AMC 10B Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Divisibility & Factors · difficulty 2 of 5 · about 2 min · position P1-10

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

A)
$6$
B)
$7$
C)
$8$
D)
$9$
E)
$10$

Answer

Key insight

Cross-multiplying gives MN = 36, so ordered pairs correspond to divisors of 36 = 2^2 * 3^2, which number (2+1)(2+1) = 9.

Solution

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 $\boxed{\textbf{(D)}\ 9}$ .

Why this works

A proportion with the same number on both diagonals hides a product equation $MN = k$ , and ordered solutions of $MN = k$ are counted by the divisor function of $k$ . The formula $\prod (e_i + 1)$ counts divisors from exponents, so no listing is needed once the factorization is known.

The trap

Counting only unordered factor pairs (5), or double counting the pair (6, 6), which is one ordered pair, not two.

Common mistakes

Counting only unordered factor pairs (5), or double counting the pair (6, 6), which is one ordered pair, not two.
Counting divisors of $6$ (4 of them) or of $12$ instead of $36$ after a cross-multiplication slip.

Techniques

Set up the equation/formula and compute; no special trick needed