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.

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

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

Answer: B
Concepts:algebraic manipulationquadratic
Difficulty rating: 1140
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+364^2 + 5^2 + 6^2 = 16 + 25 + 36 =77.= 77.

Thus, the correct answer is B.

← Problem 10#10
Full Exam

Problem 11 in Other Years