Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2020 AMC 12B Problems/Problem 6

Revision as of 07:57, 18 October 2021 by Countmath1 (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

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. We have now revealed that the condition $n \geq 9$ is insignificant and added most likely to encourage the selection of choice $\textbf{(B)}$.

~DBlack2021 (Solution Writing)

~Countmath1 (Minor edits in formatting)

~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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_6&oldid=163709"