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

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==Solution 2==
 
==Solution 2==
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We know that in order for something to be a perfect square, it has to be written as <math>x^{2}</math>. So, if we divide all of the exponents by 2, we can see which ones are perfect squares, and which ones are not. <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> So, that only leaves us with one choice, <math>\boxed{\textbf{(B) }2^{2017}}</math>.
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We know that in order for something to be a perfect square, it has to be written as <math>x^{2}</math>. So, if we divide all of the exponents by 2, we can see which ones are perfect squares, and which ones are not. <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. So, that only leaves us with one choice, <math>\boxed{\textbf{(B) }2^{2017}}</math>.
 
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{{AMC8 box|year=2016|num-b=6|num-a=8}}
 
{{AMC8 box|year=2016|num-b=6|num-a=8}}
 
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Revision as of 19:40, 1 April 2021

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 $\boxed{\textbf{(B) }2^{2017}}$.

Solution 2

We know that in order for something to be a perfect square, it has to be written as $x^{2}$. So, if we divide all of the exponents by 2, we can see which ones are perfect squares, and which ones are not. $1^{2016}=(1^{1008})^{2}$, $2^{2017}=2^{\frac {2017}{2}}$, $3^{2018}=(3^{1009})^{2}$, $4^{2019}=4^{\frac {2019}{2}}$, $5^{2020}=(5^{1010})^{2}$. 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 ${4^{2019}=4^{2018} \cdot 4}$. this is also a perfect square because the exponent $2018$ is even, and the base $4$ is also a perfect square, thus ${4^{2019}}$ is a perfect square. So, that only leaves us with one choice, $\boxed{\textbf{(B) }2^{2017}}$. -fn106068

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

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