Difference between revisions of "2007 AMC 8 Problems/Problem 19"

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Revision as of 00:29, 5 July 2013

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 $2x+1<100$. $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$. $2x+1$ 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 $\boxed{\mathrm{(C)} 79}$

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

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