2014 AMC 10A Problems/Problem 8

Problem

Which of the following number 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 $\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2\cdot(n+1)}{2}=(n!)^2\cdot\dfrac{n+1}{2}$ . Therefore, the product will only be a perfect square if the second term is a perfect square. The only answer for which the previous is true is $\dfrac{17!18!}{2}=(17!)^2\cdot9$ .

Solution 2

Notice that $17!18!=17!(17!\times 18)=(17!)^2\times 18$ . So $\dfrac{(17!)^2\times 18}{2}=(17!)^2\times 9=(17!)^2\times 3^2$ . Therefore, it is a perfect square. None of the other choices can be factored this way.

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.

Page

Toolbox

Search

2014 AMC 10A Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

See Also