2002 AMC 10B Problems/Problem 11

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 $4^2+5^2+6^2=77\Longrightarrow\boxed{\mathrm{ (B) \ }}$ .

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2002 AMC 10B Problems/Problem 11

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