2007 AMC 8 Problems/Problem 19

Problem

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?

$\mathrm{(A)}\ 2 \qquad \mathrm{(B)}\ 64 \qquad \mathrm{(C)}\ 79 \qquad \mathrm{(D)}\ 96 \qquad \mathrm{(E)}\ 131$

Solution 1

Let the smaller of the two numbers be $x$ . Then, the problem states that $(x+1)+x<100$ . $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$ . $2x+1$ is obviously odd, so only answer choices C and E need to be considered.

$2x+1=131$ contradicts the fact that $2x+1<100$ , so the answer is $\boxed{\mathrm{(C)} 79}$

Solution 2

Since for two consecutive numbers $a$ and $b$ , the difference between their squares are $a^2-b^2=(a+b)(a-b)$ , which equals to $a+b$ , because $a$ and $b$ are consecutive. And because they are consecutive, one number must be even, and the other odd. Since the sum of an even and an odd number is always odd, and that the sum of $a$ and $b$ is less than 100, you can eliminate all answers expect for $\boxed{\mathrm{(C)} 79}$ .

Video Solution

https://youtu.be/8FGl2vtO1vs Soo, DRMS, NM

Video Solution by WhyMath

https://youtu.be/BrEqmDq82rw

~savannahsolver

2007 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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