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Difference between revisions of "2014 AMC 10A Problems/Problem 8"

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<math> \textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 </math>
 
<math> \textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 </math>
  
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== Solution 1==
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== Solution==
  
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Note that
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Note that for all positive <math>n</math>, we have
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<math>\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2\cdot(n+1)}{2}=(n!)^2\cdot\dfrac{n+1}{2}</math>. Therefore, the product will only be a perfect square if the second term is a perfect square.
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<cmath>\dfrac{n!(n+1)!}{2}</cmath>
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The only answer for which the previous is true is <math>\dfrac{17!18!}{2}=(17!)^2\cdot9</math>.
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<cmath>\implies\dfrac{(n!)^2\cdot(n+1)}{2}</cmath>
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<cmath>\implies (n!)^2\cdot\dfrac{n+1}{2}</cmath>
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We must find a value of <math>n</math> such that <math>(n!)^2\cdot\dfrac{n+1}{2}</math> is a perfect square. Since <math>(n!)^2</math> is a perfect square, we must also have <math>\frac{n+1}{2}</math> be a perfect square.
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In order for <math>\frac{n+1}{2}</math> to be a perfect square, <math>n+1</math> must be twice a perfect square. From the answer choices, <math>n+1=18</math> works, thus, <math>n=17</math> and our desired answer is <math>\boxed{\textbf{(D)}\ \frac{17!18!}{2}}</math>
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(Solution/revised by bestwillcui1)
  
 
==See Also==
 
==See Also==

Revision as of 11:16, 9 February 2014

Problem

Which of the following number is a perfect square?

$\textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2$

Solution

Note that for all positive $n$, we have \[\dfrac{n!(n+1)!}{2}\] \[\implies\dfrac{(n!)^2\cdot(n+1)}{2}\] \[\implies (n!)^2\cdot\dfrac{n+1}{2}\]

We must find a value of $n$ such that $(n!)^2\cdot\dfrac{n+1}{2}$ is a perfect square. Since $(n!)^2$ is a perfect square, we must also have $\frac{n+1}{2}$ be a perfect square.

In order for $\frac{n+1}{2}$ to be a perfect square, $n+1$ must be twice a perfect square. From the answer choices, $n+1=18$ works, thus, $n=17$ and our desired answer is $\boxed{\textbf{(D)}\ \frac{17!18!}{2}}$

(Solution/revised by bestwillcui1)

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 10 Problems and Solutions

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