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Difference between revisions of "2002 AMC 10B Problems/Problem 11"

(New page: == Problem == The product of three consecutive positive integers is <math>8</math> times their sum. What is the sum of the squares? <math> \mathrm{(A) \ } 50\qquad \mathrm{(B) \ } 77\qqu...)
(No difference)

Revision as of 03:08, 27 December 2008

Problem

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

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


Solution

Let the three consecutive positive integers be $a-1$, $a$, and $a+1$. So, $a(a-1)(a+1)=a^3-a=24a$. Rearranging and factoring, $a(a+5)(a-5)=0$, so $a=5$. Hence, the sum of the squares is $4^2+5^2+6^2=77\Longrightarrow\mathrm{ (B) \ }$

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