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2002 AMC 12B Problems/Problem 7

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Problem

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

Solution

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)}.

See also

2002 AMC 12B (Problems • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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
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Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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