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

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(Solution)
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==Solution==
 
==Solution==
<cmath>\frac{(n+2)!-(n+1)!}{n!}</cmath> can be simplified by common denominator n!.
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We first expand the expression:
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>
This expression can be shown as <cmath>(n+1)(n+2-1) = (n+1)^2</cmath>
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which proves that the answer is <math>\textbf{(D)}</math>.
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We can now divide out a common factor of <math>n!</math> from each term of this expression:
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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+1)(n+2-1) = (n+1)^2</cmath>
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which proves that the answer is <math>\boxed{\textbf{(D) a perfect square}}</math>.

Revision as of 19:35, 7 February 2020

Problem 6

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

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 $\boxed{\textbf{(D) a perfect square}}$.