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> | ||
| − | ==Solution== | + | ==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>. | 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>. | ||
{{AMC8 box|year=2016|num-b=6|num-a=8}} | {{AMC8 box|year=2016|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | |||
| + | ==Solution 2== | ||
| + | We know that in order for something to be a perfect | ||
Revision as of 17:32, 1 April 2021
Problem
Which of the following numbers is not a perfect square?
Solution 1
Our answer must have an odd exponent in order for it to not be a square. Because 
is a perfect square, 
is also a perfect square, so our answer is 
.
| 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. 
Solution 2
We know that in order for something to be a perfect
