2007 AMC 8 Problem 19

Attempt Problem 19 of the 2007 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

Pick two consecutive positive integers whose sum is less than 100.100. Square both of those integers and then find the difference of the squares. Which of the following could be the difference?
22
6464
7979
9696
131131
Answer: C
Concepts:difference of squaresparity
Small hint:
Let the consecutive integers be xx and x+1x+1.
Big hint:
The difference of their squares simplifies to the sum of the two integers.
Solution:
Let the consecutive positive integers be xx and x+1x+1. Their sum is 2x+12x+1, which is less than 100100.
The difference of their squares is (x+1)2−x2=(x+1+x)(x+1−x)=2x+1. \begin{gathered} (x+1)^2-x^2 \\ = (x+1+x)(x+1-x) \\ = 2x+1. \end{gathered}
So the difference must be an odd number less than 100100. Among the choices, 7979 is the only possibility, and it occurs for 3939 and 4040.
Thus, C is the correct answer.
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