Difference between revisions of "2020 AMC 12B Problems/Problem 6"

m (Solution 2)
m (Problem)
(2 intermediate revisions by the same user not shown)
Line 4: Line 4:
 
<cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following?
 
<cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following?
  
<math>\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}</math>
+
<math>\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}</math>
  
 
==Solution 1==
 
==Solution 1==
Line 23: Line 23:
 
This leaves <math>\boxed{\textbf{(D) } \text{a perfect square}}</math> as the only answer choice left.
 
This leaves <math>\boxed{\textbf{(D) } \text{a perfect square}}</math> as the only answer choice left.
  
This solution does not consider the condition <math>n \geq 9.</math> The reason is that, with further testing it becomes clear that for all <math>n,</math> we get <cmath>(n+2)(n+1)-(n+1) = (n+1)^{2},</cmath> as proved in Solution 1. The condition <math>n \geq 9</math> was added most likely to encourage picking <math>\textbf{(B)}</math> and discourage substituting smaller values of <math>n.</math>  
+
This solution does not consider the condition <math>n \geq 9.</math> The reason is that, with further testing it becomes clear that for all <math>n,</math> we get <cmath>(n+2)(n+1)-(n+1) = (n+1)^{2},</cmath> as proved in Solution 1. The condition <math>n \geq 9</math> was added most likely to encourage picking <math>\textbf{(B)}</math> and discourage substituting smaller values into <math>n.</math>  
  
 
~DBlack2021 (Solution)
 
~DBlack2021 (Solution)

Revision as of 01:03, 28 April 2022

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

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

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