2002 AMC 12B Problem 7

Below is the professionally curated solution for Problem 7 of the 2002 AMC 12B, 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.

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Concepts:quadraticalgebraic manipulation

Difficulty rating: 1190

7.

The product of three consecutive positive integers is 88 times their sum. What is the sum of their squares?

5050

7777

110110

149149

194194

Solution:

Let the integers be n1,n-1, n,n, n+1.n+1. Then (n1)n(n+1)=83n,(n-1)n(n+1)=8\cdot3n, so n21=24n^2-1=24 and n=5.n=5.

The three integers 4,4, 5,5, 66 have squares summing to 16+25+36=77.16+25+36=77.

Thus, the correct answer is B.

Problem 7 in Other Years