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

Latest revision as of 22:02, 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

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