# 2020 AMC 12B Problems/Problem 6

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## Problem

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

$\textbf{(A) } \text{a multiple of 4} \qquad \textbf{(B) } \text{a multiple of 10} \qquad \textbf{(C) } \text{a prime number} \qquad \textbf{(D) } \text{a perfect square} \qquad \textbf{(E) } \text{a perfect cube}$

## Solution 1

We first expand the expression: $$\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}.$$ We can now divide out a common factor of $n!$ from each term of the numerator: $$(n+2)(n+1)-(n+1).$$ Factoring out $(n+1),$ we get $$[(n+2)-1](n+1) = (n+1)^2,$$ which proves that the answer is $\boxed{\textbf{(D) } \text{a perfect square}}.$

## Solution 2

In the numerator, we factor out an $n!$ to get $$\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1).$$ Now, without loss of generality, test values of $n$ until only one answer choice is left valid:

• $n = 1 \implies (3)(2) - (2) = 4,$ knocking out $\textbf{(B)},\textbf{(C)},$ and $\textbf{(E)}.$
• $n = 2 \implies (4)(3) - (3) = 9,$ knocking out $\textbf{(A)}.$

This leaves $\boxed{\textbf{(D) } \text{a perfect square}}$ as the only answer choice left.

This solution does not consider the condition $n \geq 9.$ The reason is that, with further testing it becomes clear that for all $n,$ we get $$(n+2)(n+1)-(n+1) = (n+1)^{2},$$ as proved in Solution 1. The condition $n \geq 9$ was added most likely to encourage picking $\textbf{(B)}$ and discourage substituting smaller values into $n.$

~DBlack2021 (Solution)

~MRENTHUSIASM (Edits in Logic)

~Countmath1 (Minor Edits in Formatting)

~ pi_is_3.14

## Video Solution

~IceMatrix

 2020 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions