2020 AMC 12B Problems/Problem 6

Revision as of 22:54, 8 June 2021 by MRENTHUSIASM (talk | contribs) (Solution 2)

Problem

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

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

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 this expression:

\[(n+2)(n+1)-(n+1)\]

Factoring out $(n+1)$, we get \[(n+1)(n+2-1) = (n+1)^2\]

which proves that the answer is $\boxed{\textbf{(D)} \text{ a perfect square}}$.

Solution 2

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. We have now revealed that the condition $n \geq 9$ is insignificant.

~DBlack2021 (Solution Writing)

~MRENTHUSIASM (Edits in Logic)

Video Solution

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

~ pi_is_3.14

Video Solution

https://youtu.be/6ujfjGLzVoE

~IceMatrix

See Also

2020 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 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

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