Difference between revisions of "2017 AMC 12A Problems/Problem 17"

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This is real iff <math>\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n</math> is even<math>)</math>. Thus, the answer is the number of even <math>0\leq n<24</math> which is <math>\boxed{(D)=\ 12}</math>.
 
This is real iff <math>\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n</math> is even<math>)</math>. Thus, the answer is the number of even <math>0\leq n<24</math> which is <math>\boxed{(D)=\ 12}</math>.
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==Solution 2==
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<math>z = \sqrt[24]{1} = 1^{\frac{1}{24}}</math>
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By [[Euler's identity]], <math>1 = e^{0 \cdot i} = cos (2k\pi) + i sin(2k\pi)</math>, where <math>k</math> is an integer.
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Using [[De Moivre's Theorem]], we have <math>z = 1^{\frac{1}{24}} = {cos (\frac{k\pi}{12}) + i sin (\frac{k\pi}{12})}</math>, where <math>0 \leq k<24</math> and <math>k</math> is an integer.
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Using De Moivre's Theorem again, we have that <math>z^6 = {cos (\frac{k\pi}{2}) + i sin (\frac{k\pi}{2})}</math>
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For <math>z^6</math> to be real, <math>sin(\frac{k\pi}{2})</math> has be equal <math>0</math> to negate the imaginary component. This occurs whenever <math>\frac{k\pi}{2}</math> is a multiple of <math>\pi</math>. This occurs whenever <math>k</math> is even. There are exactly <math>\boxed{12}</math> even values of <math>k</math> on the interval <math>0 \leq k<24</math>, so the answer is <math>\boxed{(D)}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2017|ab=A|num-b=16|num-a=18}}
 
{{AMC12 box|year=2017|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:06, 9 February 2017

Problem

There are $24$ different complex numbers $z$ such that $z^{24}=1$. For how many of these is $z^6$ a real number?

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24$

Solution

Note that these $z$ such that $z^{24}=1$ are $e^{\frac{ni\pi}{12}}$ for integer $0\leq n<24$. So

$z^6=e^{\frac{ni\pi}{2}}$

This is real iff $\frac{n}{2}\in \mathbb{Z} \Leftrightarrow (n$ is even$)$. Thus, the answer is the number of even $0\leq n<24$ which is $\boxed{(D)=\ 12}$.


Solution 2

$z = \sqrt[24]{1} = 1^{\frac{1}{24}}$

By Euler's identity, $1 = e^{0 \cdot i} = cos (2k\pi) + i sin(2k\pi)$, where $k$ is an integer.

Using De Moivre's Theorem, we have $z = 1^{\frac{1}{24}} = {cos (\frac{k\pi}{12}) + i sin (\frac{k\pi}{12})}$, where $0 \leq k<24$ and $k$ is an integer.

Using De Moivre's Theorem again, we have that $z^6 = {cos (\frac{k\pi}{2}) + i sin (\frac{k\pi}{2})}$

For $z^6$ to be real, $sin(\frac{k\pi}{2})$ has be equal $0$ to negate the imaginary component. This occurs whenever $\frac{k\pi}{2}$ is a multiple of $\pi$. This occurs whenever $k$ is even. There are exactly $\boxed{12}$ even values of $k$ on the interval $0 \leq k<24$, so the answer is $\boxed{(D)}$.

See Also

2017 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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 AMC 12 Problems and Solutions

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