AMC 8 · 2007 · #19

Grade 6 number-theory

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

paritylinear-equations-one-varperfect-squares convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: paritylinear-equations-one-var

📏 Medium solution 💡 2 insights

Problem

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?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

131

AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

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.

1STEP 1

Shrink the problem (Tool #9): compute the difference of squares for small pairs — 3, 5, 7, 9.

2² - 1² = 4 - 1 = 3 3² - 2² = 9 - 4 = 5 4² - 3² = 16 - 9 = 7 5² - 4² = 25 - 16 = 9

2STEP 2

Spot the pattern: those odd results are exactly the pair's own sum, so the difference of squares equals 2n + 1.

(n+1)² - n² = n + (n+1) = 2n + 1

3STEP 3

Read off two filters: the difference 2n + 1 is always odd, and it stays under 100.

difference = 2n + 1 is odd, and 2n + 1 ＜ 100

4STEP 4

Apply both filters (Tool #3): 2, 64, 96 are even; 131 tops 99 — only 79 survives, odd and below 100.

(A) 2 even ✗ (B) 64 even ✗ (C) 79 odd and 79 ＜ 100 ✓ (D) 96 even ✗ (E) 131 ＞ 99 ✗

5STEP 5

Confirm it's reachable: 2n + 1 = 79 gives n = 39, and 40² - 39² = 79. The answer is (C).

2n + 1 = 79 → n = 39 40² - 39² = 1600 - 1521 = 79 → (C)

Answer

Sanity-check with the difference-of-squares identity, which the pattern matches: (n+1)² - n² = (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.

💡Key takeaway

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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