AMC 10 · 2012 · #10
Grade 7 algebraHow many ordered pairs of positive integers (M,N) satisfy the equation 6M=N6?
Pick an answer.
AMC 10 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Count the ordered pairs of positive integers $(M,N)$ that make the proportion $\frac{M}{6}=\frac{6}{N}$ true. Because the pairs are ordered, $(M,N)$ and $(N,M)$ count as different unless they happen to be equal.
Givens: The equation $\frac{M}{6}=\frac{6}{N}$; $M$ and $N$ must both be positive integers; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Unknowns: How many ordered pairs $(M,N)$ satisfy the equation
Understand
Restated: Count the ordered pairs of positive integers $(M,N)$ that make the proportion $\frac{M}{6}=\frac{6}{N}$ true. Because the pairs are ordered, $(M,N)$ and $(N,M)$ count as different unless they happen to be equal.
Givens: The equation $\frac{M}{6}=\frac{6}{N}$; $M$ and $N$ must both be positive integers; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
The proportion hides a simpler equation, so Tool #4 (Introduce a Variable) is used first to cross-multiply $\frac{M}{6}=\frac{6}{N}$ into $MN=36$. Once the problem becomes "how many positive-integer factor pairs give $36$," Tool #2 (Make a Systematic List) walks the divisors of $36$ in order so none are missed or double-counted. Tool #3 (Eliminate Possibilities) guards against the classic trap of counting unordered pairs (which would give $5$) instead of ordered ones.
Execute — Answer: D
7.RP.A.2 Step 1 Cross-multiply the proportion
- Two equal fractions can be cross-multiplied: $\frac{M}{6}=\frac{6}{N}$ means $M\cdot N = 6\cdot 6$, so $MN=36$.
- The whole question is now "how many ordered pairs of positive integers multiply to $36$?"
💡 Equal ratios always have equal cross products, which turns a fraction equation into a clean product.
6.EE.B.6 Step 2 See it as factor pairs of 36
- For each positive integer $M$ that divides $36$, the partner $N=\frac{36}{M}$ is forced and is automatically a positive integer.
- So every ordered pair $(M,N)$ corresponds to exactly one divisor $M$ of $36$: count the divisors and you count the pairs.
💡 Fixing $M$ pins down $N$, so pairs and divisors match one-for-one.
4.OA.B.4 Step 3 List the divisors of 36 in order
- Walk the divisors of $36$ from smallest to largest: $1,2,3,4,6,9,12,18,36$.
- That is $9$ divisors, giving the $9$ ordered pairs $(1,36),(2,18),(3,12),(4,9),(6,6),(9,4),(12,3),(18,2),(36,1)$.
- Because order matters, $(1,36)$ and $(36,1)$ both count; the pair $(6,6)$ is the only self-paired one.
- So there are $9$ ordered pairs, and the answer is (D).
💡 Listing divisors in order guarantees you catch every factor pair exactly once.
7.RP.A.2 Two equal fractions can be cross-multiplied: $\frac{M}{6}=\frac{6}{N}$ means $M\ 6.EE.B.6 For each positive integer $M$ that divides $36$, the partner $N=\frac{36}{M}$ is 4.OA.B.4 Walk the divisors of $36$ from smallest to largest: $1,2,3,4,6,9,12,18,36$. That Review
Reasonableness: The count of ordered pairs equals the number of divisors of $36$. Using prime factorization, $36=2^2\cdot 3^2$, so the divisor count is $(2+1)(2+1)=9$ — matching the hand list. If you mistakenly counted unordered pairs you would get $5$ (the four flip-pairs plus $(6,6)$), which is not even an answer choice, confirming that ordered counting is intended and $9$ is right.
Alternative: Skip the listing and use the divisor-count formula directly: write $36=2^2\cdot 3^2$, add one to each exponent, and multiply: $(2+1)(2+1)=9$. Since each divisor $M$ gives one ordered pair, the answer is $9$, choice (D).
CCSS standards used (min grade 7)
7.RP.A.2Recognize and represent proportional relationships between quantities (Cross-multiplying the proportion $\frac{M}{6}=\frac{6}{N}$ into the equation $MN=36$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Treating $M$ and $N$ as unknowns linked by one equation, so choosing $M$ forces $N=\frac{36}{M}$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing every divisor of $36$ to count the ordered factor pairs.)
⭐ Cross-multiply the proportion to get $MN=36$, then count the divisors of $36$ — each divisor is one ordered pair.
⭐ Cross-multiply the proportion to get $MN=36$, then count the divisors of $36$ — each divisor is one ordered pair.
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