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2020 AMC 12B Problem 6

Problem 6 of 25EasierAlgebraNumber TheoryCounting & Probability

For all integers n≥9,n \ge 9, the value of (n+2)!−(n+1)!n!\frac{(n + 2)! - (n + 1)!}{n!} is always which of the following?

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Solution

Factor (n+1)!(n + 1)! from the numerator: (n+2)!−(n+1)!=(n+1)! [(n+2)−1]=(n+1)! (n+1). \begin{gathered} (n + 2)! - (n + 1)! \\ {}= (n + 1)!\,\big[(n + 2) - 1\big] \\ {}= (n + 1)!\,(n + 1). \end{gathered} Dividing by n!n! leaves (n+1)! (n+1)n!\dfrac{(n + 1)!\,(n + 1)}{n!} =(n+1)(n+1)= (n + 1)(n + 1) =(n+1)2,= (n + 1)^2, which is always a perfect square. Thus, the correct answer is D.

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Concepts: factorial · factoring · perfect square

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