2007 AMC 8 Problems/Problem 19

Revision as of 22:16, 11 December 2021 by Mathwinner1 (talk | contribs) (Solution)

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.

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

Video Solution by WhyMath

https://youtu.be/BrEqmDq82rw

~savannahsolver

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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