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2007 AMC 8

Problem 19

Problem 19 · 2007 AMC 8 Easy
Algebra & Patterns difference-of-squares

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?

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Answer: C — 79.
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Hint 1 of 2
Try it with real numbers first: 42 − 32 = 7, and 3 + 4 = 7. The difference of the squares is just the sum of the two integers — which is always odd. So two of the choices are dead on arrival.
Still stuck? Show hint 2 →
Hint 2 of 2
Difference of squares: p2 − q2 = (p+q)(p−q); for consecutive integers the second factor is 1, so it collapses to p+q.
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Approach: factor the difference, then screen by parity
  1. Let the integers be x and x+1. Then (x+1)2 − x2 = (2x+1)(1) = 2x+1 — exactly the sum of the two numbers.
  2. So the difference is odd (it's 2x+1) and, since the sum is under 100, less than 100.
  3. Scan the choices: 2 and 64 are even (out); 131 is over 100 (out). Only 79 is odd and below 100.
  4. Big idea: the gap between consecutive squares grows by odd numbers (1, 3, 5, 7, …) — so every consecutive-square difference is odd. That parity test killed three options instantly.
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