2002 AMC 10B Problem 11

Below is the professionally curated solution for Problem 11 of the 2002 AMC 10B, 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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Concepts:algebraic manipulationquadratic

Difficulty rating: 1140

11.

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. Their product is n(n21)n(n^2 - 1) and their sum is 3n,3n, so n(n21)=8(3n)=24n.n(n^2 - 1) = 8(3n) = 24n.

Since n0,n \ne 0, we get n21=24,n^2 - 1 = 24, so n2=25n^2 = 25 and n=5.n = 5.

The three integers are 4,5,6,4, 5, 6, and 42+52+62=16+25+36=77.4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77.

Thus, the correct answer is B.

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