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2002 AMC 12B Problem 7

Problem 7 of 25EasierAlgebra

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

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Solution

Let the integers be n−1,n-1, n,n, n+1.n+1. Then (n−1)n(n+1)=8⋅3n,(n-1)n(n+1)=8\cdot3n, so n2−1=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.

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Concepts: quadratic · algebraic manipulation

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.