AMC 8 · 2007 · #19
Grade 6 number-theoryProblem
Pick two consecutive positive integers whose sum is less than . Square both
of those integers and then find the difference of the squares. Which of the
following could be the difference?
Pick an answer.
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) $79$.