2002 AMC 12B Problem 7 — Solution | LIVE by Po-Shen Loh

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2002 AMC 12B Problem 7

Attempt Problem 7 of the 2002 AMC 12B 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 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

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:quadratic algebraic manipulation

Difficulty rating: 1190

Solution:

Let the integers be $n-1,$ $n,$ $n+1.$ Then $(n-1)n(n+1)=8\cdot3n,$ so $n^2-1=24$ and $n=5.$

The three integers $4,$ $5,$ $6$ have squares summing to $16+25+36=77.$

Thus, the correct answer is B.

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