Difference between revisions of "2002 AMC 12B Problems/Problem 7"

Revision as of 21:13, 6 June 2011

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?

$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$

Let the three consecutive integers be $x-1, x, x+1$ ; then $(x-1)(x)(x+1) = x(x^2 - 1) = 8(x-1 + x + x+1) = 24x$

Since $x \neq 0$ , we have $x^2 = 25$ , with the positive solution being $x = 5$ . Then $4^2 + 5^2 + 6^2 = 77\ \mathrm{(B)}$ .

2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by [[2002 AMC 12B Problems/Problem {{{num-a}}}\|Problem {{{num-a}}}]]
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All AMC 12 Problems and Solutions

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 == See also ==
-{{AMC12 box|year=2002|ab=B|num-b=6|num-a=8}}
+{{AMC12 box|year=2002|ab=B|num-b=6|num-876897689798=8}}
 [[Category:Introductory Algebra Problems]]