 # What are we learning ?

Competency in Focus: Perfect square numbers  This problem from American Mathematics contest (AMC 10A, 2014) is based on the concept that when a number is a perfect square .

# First look at the knowledge graph. # Next understand the problem

Which of the following numbers is a perfect square? $\textbf{(A)}\ \dfrac{14!15!}2\qquad\textbf{(B)}\ \dfrac{15!16!}2\qquad\textbf{(C)}\ \dfrac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2$
##### Source of the problem
American Mathematical Contest 2014, AMC 10A  Problem 8
##### Key Competency
This number theory problem is based on the concept that when a number is a perfect square
5/10
##### Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

Do you really need a hint? Try it first!
First of all look at the examples , see that   for all positive $n$, we have $$\dfrac{n!(n+1)!}{2}$$.Now what we have to do with this ?
Now we have to find which member has what uniform numbers from the given  conversation .
After some simple manipulations , we have $$\dfrac{n!(n+1)!}{2}$$ $$\implies\dfrac{(n!)^2\cdot(n+1)}{2}$$ $$\implies (n!)^2\cdot\dfrac{n+1}{2}$$ . Thus now the problem reduces to  finding  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.

Now check the options and see for what value of n , $n+1$ must be twice a perfect square. $n+1=18$ works, thus, $n=17$ and our desired answer is $\boxed{\textbf{(D)}\ \frac{17!18!}{2}}$

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