2002 AMC 10B Problem 11

Problem 11 of 25IntermediateAlgebra

The product of three consecutive positive integers is

8

times their sum. What is the sum of their squares?

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Solution

Let the integers be

n - 1,

n,

n + 1.

Their product is

n(n^2 - 1)

and their sum is

3n,

n(n^2 - 1) = 8(3n) = 24n.

Since

n \ne 0,

we get

n^2 - 1 = 24,

n^2 = 25

and

n = 5.

The three integers are

4,

5,

and

6,

and

4^2 + 5^2 + 6^2 = 16 + 25 + 36

= 77.

Thus, the correct answer is B.