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

Revision as of 00:39, 24 December 2022

Problem

Which of the following numbers is not a perfect square?

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

Video Solution

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

https://youtu.be/prtDHdc12cs

~savannahsolver

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 can be written as $x^{2}$ for $x \in R$ . 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)
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.

@@ Line 17: / Line 17: @@
 ==Solution 2==
-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>.
+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 R</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>.
 -fn106068

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