AMC 8 · 2007 · #19
Grade 6 number-theoryProblem
Pick two consecutive positive integers whose sum is less than 100. Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
Pick an answer.
AMC 8 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: Pick any two consecutive positive integers whose sum is less than $100$. Square each and subtract the smaller square from the larger. Which of the five answer choices could be that difference?
Givens: Two consecutive positive integers, call them $n$ and $n+1$; Their sum satisfies $n + (n+1) < 100$; The quantity in question is $(n+1)^2 - n^2$; Answer choices: (A) $2$, (B) $64$, (C) $79$, (D) $96$, (E) $131$
Unknowns: Which of the five values can actually equal $(n+1)^2 - n^2$ for some allowed $n$
Understand
Restated: Pick any two consecutive positive integers whose sum is less than $100$. Square each and subtract the smaller square from the larger. Which of the five answer choices could be that difference?
Givens: Two consecutive positive integers, call them $n$ and $n+1$; Their sum satisfies $n + (n+1) < 100$; The quantity in question is $(n+1)^2 - n^2$; Answer choices: (A) $2$, (B) $64$, (C) $79$, (D) $96$, (E) $131$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities
Squaring then subtracting sounds heavy, so start with Tool #9 (Easier Related Problem): try the smallest consecutive pairs and just compute. The numbers fall into a clean pattern, which is Tool #5 (Look for a Pattern) — the difference of squares of consecutive integers equals the sum of those integers, so it is always odd and at most $99$. Tool #3 (Eliminate Possibilities) then crosses out every choice that fails the odd-and-less-than-$100$ filter. We do not need Tool #13 (Algebra) to set up an equation; the pattern from a few small cases is doing all the work.
Execute — Answer: C
6.EE.A.1 Step 1 - Try small consecutive pairs and compute the difference of squares directly.
- Tool #9 says to shrink the problem until the structure shows up.
💡 Squaring sounds intimidating, but tiny pairs make the arithmetic trivial and show the structure right away.
6.EE.A.3 Step 2 - Spot the pattern.
- The differences $3, 5, 7, 9, \dots$ are exactly the odd numbers, and each one equals the sum of the pair itself: $1+2=3$, $2+3=5$, $3+4=7$, $4+5=9$.
- So the difference of squares of two consecutive integers equals their sum.
💡 Whatever you put in, the answer always comes out as twice $n$ plus one — an odd number, and the same as $n + (n+1)$.
6.NS.B.4 Step 3 - Read off two filters from the pattern.
- First, $2n + 1$ is always odd.
- Second, the problem says the sum $2n + 1$ is less than $100$.
- So the difference must be an odd number less than $100$.
💡 $2n$ is even, so adding $1$ always lands on odd. And $2n+1 < 100$ comes straight from the problem's "sum less than $100$" condition.
6.EE.B.5 Step 4 - Apply the two filters to the answer choices (Tool #3).
- $2$, $64$, and $96$ are even — out.
- $131$ is odd but exceeds $99$ — out.
- Only $79$ is odd and below $100$, so it survives.
💡 Two quick tests (odd? under $100$?) crush four of the five choices in one pass.
6.EE.B.7 Step 5 - Confirm $79$ is actually reachable by finding the pair.
- Set $2n + 1 = 79$, so $n = 39$.
- Pair $39, 40$: sum $79 < 100$, and $40^2 - 39^2 = 1600 - 1521 = 79$.
- The answer is (C).
💡 Showing the actual pair $(39, 40)$ proves $79$ is achievable, not just "not yet eliminated".
6.EE.A.1 Try small consecutive pairs and compute the difference of squares directly. Tool 6.EE.A.3 Spot the pattern. The differences $3, 5, 7, 9, \dots$ are exactly the odd number 6.NS.B.4 Read off two filters from the pattern. First, $2n + 1$ is always odd. Second, th 6.EE.B.5 Apply the two filters to the answer choices (Tool #3). $2$, $64$, and $96$ are e 6.EE.B.7 Confirm $79$ is actually reachable by finding the pair. Set $2n + 1 = 79$, so $n Review
Reasonableness: Sanity-check with the difference-of-squares identity, which the pattern matches: $(n+1)^2 - n^2 = (n+1+n)(n+1-n) = (2n+1)(1) = 2n+1$. So the difference is exactly the sum of the two integers, which is given to be less than $100$ and is always odd because one of two consecutive integers is even and the other is odd. Among the choices, only $79$ is an odd number under $100$, and the pair $39, 40$ produces it. The answer (C) is consistent.
Alternative: Tool #3 alone (Eliminate Possibilities), no pattern needed: $(n+1)^2 - n^2$ factors as $(n+1+n)(n+1-n) = 2n+1$, which is the sum. The problem already restricts the sum to be less than $100$, and consecutive integers (one even, one odd) sum to an odd number. That immediately rules out $2$, $64$, $96$ (even) and $131$ (too big), leaving (C) $79$.
CCSS standards used (min grade 6)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Computing $2^2 - 1^2$, $3^2 - 2^2$, and so on for the small-case pairs that reveal the pattern.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rewriting $(n+1)^2 - n^2$ as the equivalent expression $2n + 1$, which is the sum of the two consecutive integers.)6.NS.B.4Find common factors and multiples; recognize even and odd numbers (Concluding that $2n + 1$ is always odd, which eliminates the three even answer choices.)6.EE.B.5Understand solving an equation or inequality as a process of answering which values make the statement true (Testing each answer choice against the two conditions (odd, less than $100$) to find which value the difference can actually equal.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Solving $2n + 1 = 79$ to recover the witness pair $n = 39$, confirming $79$ is achievable.)
⭐ The difference of squares of two consecutive integers is just their sum — always odd, and here below $100$. That one fact knocks out four choices and lands on (C) $79$.
⭐ The difference of squares of two consecutive integers is just their sum — always odd, and here below $100$. That one fact knocks out four choices and lands on (C) $79$.
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