Difference between revisions of "2016 AMC 8 Problems/Problem 7"
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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>, 3^2018, 4^2019, 5^2020 | + | 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}</math> |
Revision as of 18:18, 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 square, it has to be written as 
. So, if we divide all of the exponents by 2, we can see which ones are perfect squares, and which ones are not. 
, 
, $3^{2018}=(3^{1009}^{2}$ (Error compiling LaTeX. Unknown error_msg), 
, 
