2020 AMC 12B Problems/Problem 6

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)

Video Solution by Omega Learn

https://youtu.be/ba6w1OhXqOQ?t=2234

~ pi_is_3.14

Video Solution

https://youtu.be/6ujfjGLzVoE

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 12 Problems and Solutions

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