2002 AMC 10B Problem 11

Attempt Problem 11 of the 2002 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10B solutions, or check the answer key.

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11.

The product of three consecutive positive integers is

8

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

50

77

110

149

194

Answer: B

Concepts:algebraic manipulation quadratic

Difficulty rating: 1140

Small hint:

Call the integers

n - 1,

n,

n + 1

; their sum is

3n

Big hint:

The product is

(n-1)n(n+1) = n(n^2 - 1),

which equals

8\cdot 3n

Solution:

Let the integers be

n - 1,

n,

n + 1.

Their product is

n(n^2 - 1)

and their sum is

3n,

so

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

Since

n \ne 0,

we get

n^2 - 1 = 24,

so

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.