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

Latest revision as of 23:09, 26 June 2023

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

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
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.

@@ Line 15: / Line 15: @@
 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>
+==Video Solution (CREATIVE THINKING)==
+ https://youtu.be/sa9OON6KXb8
+~Education, the Study of Everything
 ==Video Solution==

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