AMC 10 · 2007 · #23
Grade 6 arithmeticHow many ordered pairs (m,n) of positive integers, with m≥n, have the property that their squares differ by 96?
Pick an answer.
AMC 10 2007 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 $(m,n)$ of positive integers with $m \ge n$ whose squares differ by $96$, that is, $m^2 - n^2 = 96$.
Givens: $m$ and $n$ are positive integers; $m \ge n$; $m^2 - n^2 = 96$; Answer choices: (A) $3$, (B) $4$, (C) $6$, (D) $9$, (E) $12$
Unknowns: How many pairs $(m,n)$ satisfy all the conditions
Understand
Restated: Count the ordered pairs $(m,n)$ of positive integers with $m \ge n$ whose squares differ by $96$, that is, $m^2 - n^2 = 96$.
Givens: $m$ and $n$ are positive integers; $m \ge n$; $m^2 - n^2 = 96$; Answer choices: (A) $3$, (B) $4$, (C) $6$, (D) $9$, (E) $12$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities
Chasing $m^2 - n^2 = 96$ by testing squares one at a time is slow. The unlocking move is Tool #4 (Introduce a Variable): a difference of squares always factors, $m^2 - n^2 = (m-n)(m+n)$, so the condition becomes a product of two whole numbers equal to $96$. Naming those two factors turns the problem into 'split $96$ into a product of two pieces.' From there Tool #2 (Make a Systematic List) counts the factor pairs, and Tool #3 (Eliminate Possibilities) throws out the pairs that would force $m$ or $n$ to be a fraction.
Execute — Answer: B
6.EE.A.3 Step 1 Factor the difference of squares
- A difference of two squares always breaks into a product: $m^2 - n^2 = (m-n)(m+n)$.
- So the condition $m^2 - n^2 = 96$ is exactly the statement that the two whole numbers $m-n$ and $m+n$ multiply to $96$.
💡 Rewriting the subtraction as a multiplication lets you use the factors of $96$ instead of hunting through squares.
6.EE.B.6 Step 2 Name the two factors
- Let $a = m - n$ and $b = m + n$.
- Then $a \cdot b = 96$, and because $n$ is a positive integer we have $a < b$ (the smaller factor is $m-n$, the larger is $m+n$).
- Going backwards, once you know $a$ and $b$ you recover $m = \tfrac{a+b}{2}$ and $n = \tfrac{b-a}{2}$.
- So every allowed factor pair $(a,b)$ gives exactly one pair $(m,n)$.
💡 Counting factor pairs of $96$ is the same as counting the pairs $(m,n)$, since each factor pair rebuilds one $(m,n)$.
2.OA.C.3 Step 3 Both factors must be even
- For $m = \tfrac{a+b}{2}$ to be a whole number, $a + b$ must be even, so $a$ and $b$ must share the same parity (both even or both odd).
- But their product is $96$, an even number, so they cannot both be odd.
- Therefore $a$ and $b$ are both even.
- Any factor pair of $96$ with one odd and one even number is impossible and gets thrown out.
💡 Splitting a sum in half only lands on a whole number when the two pieces are both even or both odd.
4.OA.B.4 Step 4 Count the even factor pairs
- Since $a$ and $b$ are both even, write $a = 2p$ and $b = 2q$.
- Then $(2p)(2q) = 96$ gives $4pq = 96$, so $pq = 24$, with $p < q$ positive integers.
- Now just list the factor pairs of $24$ with the smaller number first: $1 \times 24$, $2 \times 12$, $3 \times 8$, $4 \times 6$.
- That is $4$ pairs (and there is no equal case, since $24$ is not a perfect square).
- Each one rebuilds a valid pair: $(25,23),\ (14,10),\ (11,5),\ (10,2)$.
- So there are $4$ ordered pairs, which is choice (B).
💡 After pulling a factor of $2$ out of each side, the whole question shrinks to counting the factor pairs of $24$.
6.EE.A.3 A difference of two squares always breaks into a product: $m^2 - n^2 = (m-n)(m+n 6.EE.B.6 Let $a = m - n$ and $b = m + n$. Then $a \cdot b = 96$, and because $n$ is a pos 2.OA.C.3 For $m = \tfrac{a+b}{2}$ to be a whole number, $a + b$ must be even, so $a$ and 4.OA.B.4 Since $a$ and $b$ are both even, write $a = 2p$ and $b = 2q$. Then $(2p)(2q) = 9 Review
Reasonableness: Check each pair directly: $25^2 - 23^2 = 625 - 529 = 96$, $14^2 - 10^2 = 196 - 100 = 96$, $11^2 - 5^2 = 121 - 25 = 96$, and $10^2 - 2^2 = 100 - 4 = 96$. All four work, and no other even factor pair of $96$ was missed, so the count of $4$ is solid. It also makes sense that the answer is small: $96$ has only a handful of even-times-even splittings.
Alternative: List every factor pair of $96$ up front: $1\times96,\ 2\times48,\ 3\times32,\ 4\times24,\ 6\times16,\ 8\times12$. For $m-n$ and $m+n$ to give whole numbers, the two factors must have the same parity; the odd-times-even pairs $1\times96$ and $3\times32$ fail. The remaining four pairs $2\times48,\ 4\times24,\ 6\times16,\ 8\times12$ each give a valid $(m,n)$, so the answer is $4$.
CCSS standards used (min grade 6)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting the difference of squares $m^2 - n^2$ as the product $(m-n)(m+n)$.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the factors $a = m-n$ and $b = m+n$ and recovering $m = \tfrac{a+b}{2}$, $n = \tfrac{b-a}{2}$.)2.OA.C.3Determine whether a group of objects has an odd or even number (Arguing that $a$ and $b$ must both be even so that $\tfrac{a+b}{2}$ is a whole number.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing the factor pairs of $24$ (after the substitution $pq = 24$) to count the four solutions.)
⭐ A difference of squares $m^2 - n^2$ equals $(m-n)(m+n)$, so counting the pairs becomes counting the ways to split $96$ into two even factors.
⭐ A difference of squares $m^2 - n^2$ equals $(m-n)(m+n)$, so counting the pairs becomes counting the ways to split $96$ into two even factors.
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