Difference between revisions of "2016 AMC 8 Problems/Problem 7"
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| − | We know that 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 | + | We know that 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 must be <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}} | ||
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Revision as of 19:29, 27 November 2016
Which of the following numbers is not a perfect square?
Solution
We know that 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 must be 
.
| 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. 
