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

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.

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

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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2007 AMC 8 Problems/Problem 19

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Problem

Solution

Video Solution by WhyMath

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