Difference between revisions of "2016 AMC 8 Problems/Problem 7"
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+ | ==Problem== | ||
+ | |||
Which of the following numbers is not a perfect square? | Which of the following numbers is not a perfect square? | ||
<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>. |
+ | |||
+ | ==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>. | ||
+ | -fn106068 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2016|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 16:30, 21 April 2021
Contents
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 .
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. , , , , . 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 . this is also a perfect square because the exponent is even, and the base is also a perfect square, thus is a perfect square. So, that only leaves us with one choice, . -fn106068
See Also
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.