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2014 AMC 10A Problems/Problem 8: Difference between revisions

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==Problem==
==Problem==


Which of the following number is a perfect square?
Which of the following numbers is a perfect square?


<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==
== Solution 1==


Note that
Note that for all positive <math>n</math>, we have
<math>\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2*(n+1)}{2}=(n!)^2*\dfrac{n+1}{2}</math>. Therefore, the product will only be a perfect square if the second term is a perfect square.
<cmath>\dfrac{n!(n+1)!}{2}</cmath>
The only answer for which the previous is true is <math>\dfrac{17!18!}{2}=(17!)^2*9</math>.
<cmath>\implies\dfrac{(n!)^2\cdot(n+1)}{2}</cmath>
<cmath>\implies (n!)^2\cdot\dfrac{n+1}{2}</cmath>  


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.


==Solution 2==
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>


Notice that <math>17!18!=17!(17!\times 18)=(17!)^2\times 18</math>. So <math>\dfrac{(17!)^2\times 18}{2}=(17!)^2\times 9=(17!)^2\times 3^2</math>. Therefore, it is a perfect square. None of the other choices can be factored this way.
==Solution 2 (Variation of Solution 1)==
 
We take the first option A. \( \frac{14! \cdot 15!}{2} \). This is just equal to \( 14!^2 \cdot \frac{15}{2} \). \( \frac{15}{2} \) is 7.5, and 7.5 is not a perfect square, therefore option A is incorrect.
 
We can also rule that \( \frac{17}{2} = 8.5 \) and \( \frac{19}{2} = 9.5 \) are both not perfect squares, so option choices C and E are also incorrect.
 
If we take B. \( \frac{15! \cdot 16!}{2} \), we see that \( 15!^2 \) is a perfect square, but \( \frac{16}{2} = 8 \) is not, so B is incorrect. Our remaining option is <math>\boxed{\textbf{(D)}\ \frac{17!18!}{2}}</math>.
 
Proof:
 
<cmath>\frac{17! \cdot 18!}{2} \Rightarrow 17!^2 \cdot \frac{18}{2} \Rightarrow 17!^2 \cdot 9 \Rightarrow 17!^2 \cdot 3^2.</cmath>  
 
The product of two squares is always a perfect square.
 
~Pinotation
 
==Video Solution (CREATIVE THINKING)==
https://youtu.be/sa9OON6KXb8
 
~Education, the Study of Everything
 
==Video Solution==
https://youtu.be/uY_Xp8GtXP8
 
~savannahsolver


==See Also==
==See Also==
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{{AMC10 box|year=2014|ab=A|num-b=7|num-a=9}}
{{AMC10 box|year=2014|ab=A|num-b=7|num-a=9}}
{{MAA Notice}}
{{MAA Notice}}
[[Category: Introductory Number Theory Problems]]

Latest revision as of 21:02, 24 August 2025

Problem

Which of the following numbers 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 1

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 2 (Variation of Solution 1)

We take the first option A. \( \frac{14! \cdot 15!}{2} \). This is just equal to \( 14!^2 \cdot \frac{15}{2} \). \( \frac{15}{2} \) is 7.5, and 7.5 is not a perfect square, therefore option A is incorrect.

We can also rule that \( \frac{17}{2} = 8.5 \) and \( \frac{19}{2} = 9.5 \) are both not perfect squares, so option choices C and E are also incorrect.

If we take B. \( \frac{15! \cdot 16!}{2} \), we see that \( 15!^2 \) is a perfect square, but \( \frac{16}{2} = 8 \) is not, so B is incorrect. Our remaining option is $\boxed{\textbf{(D)}\ \frac{17!18!}{2}}$.

Proof:

\[\frac{17! \cdot 18!}{2} \Rightarrow 17!^2 \cdot \frac{18}{2} \Rightarrow 17!^2 \cdot 9 \Rightarrow 17!^2 \cdot 3^2.\]

The product of two squares is always a perfect square.

~Pinotation

Video Solution (CREATIVE THINKING)

https://youtu.be/sa9OON6KXb8

~Education, the Study of Everything

Video Solution

https://youtu.be/uY_Xp8GtXP8

~savannahsolver

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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