# 2014 AMC 10A Problems/Problem 8

## 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

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 $\boxed{\textbf{(D)}\ \frac{17!18!}{2}}$

~savannahsolver

## See Also

 2014 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.