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

(→‎Solution 2: The original solution violates the condition that n>=9. I am adding explanation why violating it is OK.)
m (→‎Solution 2)
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==Solution 2==
 
==Solution 2==
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Factor out an <math>n!</math> to get: <math>\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1)</math> Now, without loss of generality, test values of <math>n</math> until only one answer choice is left valid:
+
Factor out an <math>n!</math> to get <cmath>\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1).</cmath> Now, without loss of generality, test values of <math>n</math> until only one answer choice is left valid:
  
−
<math>n = 1 \implies (3)(2) - (2) = 4</math>, knocking out <math>\textbf{B}</math>, <math>\textbf{C}</math>, and <math>\textbf{E}</math>.
+
<math>n = 1 \implies (3)(2) - (2) = 4</math>, knocking out <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(E)}</math>.
−
<cmath> </cmath>
+
 
−
<math>n = 2 \implies (4)(3) - (3) = 9</math>, knocking out <math>\textbf{A}</math>.
+
<math>n = 2 \implies (4)(3) - (3) = 9</math>, knocking out <math>\textbf{(A)}.</math>
  
 
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>, <math>(n+2)(n+1)-(n+1) = (n+1)^{2}</math>, proved in Solution 1. We have now revealed that the condition <math>n \geq 9</math> is insignificant.
+
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. We have now revealed that the condition <math>n \geq 9</math> is insignificant.
  
 
~DBlack2021 (Solution Writing)
 
~DBlack2021 (Solution Writing)

Revision as of 22:54, 8 June 2021

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 (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

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