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

(Created page with "==Problem 6== For all integers <math>n \geq 9,</math> the value of <cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath>is always which of the following? <math>\textbf{(A) } \text{a multi...")
 
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==Problem 6==
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==Problem==
  
 
For all integers <math>n \geq 9,</math> the value of
 
For all integers <math>n \geq 9,</math> the value of
 
<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?
  
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<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>
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<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>
  
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==Solution==
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==Solution 1==
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<cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath> can be simplified by common denominator n!.
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We first expand the expression:
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Therefore, <cmath>\frac{(n+2)!-(n+1)!}{n!} = (n+2)(n+1)-(n+1)</cmath>
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<cmath>\frac{(n+2)!-(n+1)!}{n!} = \frac{(n+2)(n+1)n!-(n+1)n!}{n!}.</cmath>
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This expression can be shown as <cmath>\(n+1)(n+2-1) = (n+1)^2</cmath>
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We can now divide out a common factor of <math>n!</math> from each term of the numerator:
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which proves that the answer is $\textbf{(D)}.
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<cmath>(n+2)(n+1)-(n+1).</cmath>
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Factoring out <math>(n+1),</math> we get <cmath>[(n+2)-1](n+1) = (n+1)^2,</cmath>
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which proves that the answer is <math>\boxed{\textbf{(D) } \text{a perfect square}}.</math>
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==Solution 2==
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In the numerator, we 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:
 +
 
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* <math>n = 1 \implies (3)(2) - (2) = 4,</math> knocking out <math>\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(E)}.</math>
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* <math>n = 2 \implies (4)(3) - (3) = 9,</math> knocking out <math>\textbf{(A)}.</math>
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This leaves <math>\boxed{\textbf{(D) } \text{a perfect square}}</math> as the only answer choice left.
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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)
 +
 
 +
~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==
 +
 
 +
{{AMC12 box|year=2020|ab=B|num-b=5|num-a=7}}
 +
{{MAA Notice}}

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 (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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