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Difference between revisions of "2016 AMC 8 Problems/Problem 7"

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<math>\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}</math>
 
<math>\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}</math>
  
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==Solution 1==
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Our answer must have an odd exponent in order for it to not be a square.  Because <math>4</math> is a perfect square, <math>4^{2019}</math> is also a perfect square, so our answer is <math>\boxed{\textbf{(B) }2^{2017}}</math>.
  
 
==Video Solution==
 
==Video Solution==
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https://www.youtube.com/watch?v=BZKzpY_pH5A
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 +
https://youtu.be/ymZQgW-cL4s?si=NbFD5ZIPgtojqEaj
 +
 
 +
A solution so simple a 12-year-old made it!
 +
 
 +
~Elijahman~
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/_zfiULRR3co
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=BZKzpY_pH5A   ~David
  
 
https://youtu.be/prtDHdc12cs
 
https://youtu.be/prtDHdc12cs
  
 
~savannahsolver
 
~savannahsolver
−
 
−
==Solution 1==
 
−
Our answer must have an odd exponent in order for it to not be a square.  Because <math>4</math> is a perfect square, <math>4^{2019}</math> is also a perfect square, so our answer is <math>\boxed{\textbf{(B) }2^{2017}}</math>.
 
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==Solution 2==
 
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We know that in order for something to be a perfect square, it can be written as <math>x^{2}</math> for <math>x \in \mathbb{R}</math>. So, if we divide all of the exponents by 2, we can identify the perfect squares and find the answer by process of elimination. <math>1^{2016}=(1^{1008})^{2}</math>, <math>2^{2017}=2^{\frac {2017}{2}}</math>, <math>3^{2018}=(3^{1009})^{2}</math>, <math>4^{2019}=4^{\frac {2019}{2}}</math>, <math>5^{2020}=(5^{1010})^{2}</math>. Since we know that 4 is a perfect square itself, we know that even though the integer number is odd, the number that it becomes will be a perfect square. This is because <math>{4^{2019}=4^{2018} \cdot 4}</math>. this is also a perfect square because the exponent <math>2018</math> is even, and the base <math>4</math> is also a perfect square, thus <math>{4^{2019}}</math> is a perfect square. This leaves <math>\boxed{\textbf{(B) }2^{2017}}</math>.
 
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-fn106068
 
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minor edits by ~megaboy6679
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2016|num-b=6|num-a=8}}
 
{{AMC8 box|year=2016|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:47, 20 June 2024

Problem

Which of the following numbers is not a perfect square?

$\textbf{(A) }1^{2016}\qquad\textbf{(B) }2^{2017}\qquad\textbf{(C) }3^{2018}\qquad\textbf{(D) }4^{2019}\qquad \textbf{(E) }5^{2020}$

Solution 1

Our answer must have an odd exponent in order for it to not be a square. Because $4$ is a perfect square, $4^{2019}$ is also a perfect square, so our answer is $\boxed{\textbf{(B) }2^{2017}}$.

Video Solution

https://youtu.be/ymZQgW-cL4s?si=NbFD5ZIPgtojqEaj

A solution so simple a 12-year-old made it!

~Elijahman~

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/_zfiULRR3co

~Education, the Study of Everything

Video Solution

https://www.youtube.com/watch?v=BZKzpY_pH5A ~David

https://youtu.be/prtDHdc12cs

~savannahsolver

See Also

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 AJHSME/AMC 8 Problems and Solutions

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

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