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== | ||
+ | <math>z = \sqrt[24]{1} = 1^{\frac{1}{24}}</math> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Using De Moivre's Theorem again, we have that <math>z^6 = {cos (\frac{k\pi}{2}) + i sin (\frac{k\pi}{2})}</math> | ||
+ | |||
+ | 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
Contents
Problem
There are different complex numbers such that . For how many of these is a real number?
Solution
Note that these such that are for integer . So
This is real iff is even. Thus, the answer is the number of even which is .
Solution 2
By Euler's identity, , where is an integer.
Using De Moivre's Theorem, we have , where and is an integer.
Using De Moivre's Theorem again, we have that
For to be real, has be equal to negate the imaginary component. This occurs whenever is a multiple of . This occurs whenever is even. There are exactly even values of on the interval , so the answer is .
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
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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