Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
(→‎Video Solution (CREATIVE THINKING))
 
(10 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
  
−
Which of the following number is a perfect square?
+
Which of the following numbers is a perfect square?
  
 
<math> \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 </math>
 
<math> \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 </math>
  
−
== Solution 1==
+
== Solution ==
  
−
Note that
+
Note that for all positive <math>n</math>, we have
−
<math>\dfrac{n!(n+1)!}{2}=\dfrac{(n!)^2*(n+1)}{2}=(n!)^2*\dfrac{n+1}{2}</math>. Therefore, the product will only be a perfect square if the second term is a perfect square.
+
<cmath>\dfrac{n!(n+1)!}{2}</cmath>
−
The only answer for which the previous is true is <math>\dfrac{17!18!}{2}=(17!)^2*9</math>.
+
<cmath>\implies\dfrac{(n!)^2\cdot(n+1)}{2}</cmath>
 +
<cmath>\implies (n!)^2\cdot\dfrac{n+1}{2}</cmath>  
  
 +
We must find a value of <math>n</math> such that <math>(n!)^2\cdot\dfrac{n+1}{2}</math> is a perfect square. Since <math>(n!)^2</math> is a perfect square, we must also have <math>\frac{n+1}{2}</math> be a perfect square.
  
−
==Solution 2==
+
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>
  
−
Notice that <math>17!18!=17!(17!\times 18)=(17!)^2\times 18</math>. So <math>\dfrac{(17!)^2\times 18}{2}=(17!)^2\times 9=(17!)^2\times 3^2</math>. Therefore, it is a perfect square. None of the other choices can be factored this way.
+
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/sa9OON6KXb8
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://youtu.be/uY_Xp8GtXP8
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
Line 20: Line 30:
 
{{AMC10 box|year=2014|ab=A|num-b=7|num-a=9}}
 
{{AMC10 box|year=2014|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category: Introductory Number Theory Problems]]

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

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_8&oldid=194835"