Difference between revisions of "2017 AMC 12A Problems/Problem 17"
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<math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24 </math> | <math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 24 </math> | ||
− | ==Solution== | + | ==Solution 1== |
Note that these <math>z</math> such that <math>z^{24}=1</math> are <math>e^{\frac{ni\pi}{12}}</math> for integer <math>0\leq n<24</math>. So | Note that these <math>z</math> such that <math>z^{24}=1</math> are <math>e^{\frac{ni\pi}{12}}</math> for integer <math>0\leq n<24</math>. So | ||
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<math>z^6=e^{\frac{ni\pi}{2}}</math> | <math>z^6=e^{\frac{ni\pi}{2}}</math> | ||
− | This is real | + | This is real if <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>. |
− | |||
==Solution 2== | ==Solution 2== | ||
<math>z = \sqrt[24]{1} = 1^{\frac{1}{24}}</math> | <math>z = \sqrt[24]{1} = 1^{\frac{1}{24}}</math> | ||
− | By [[Euler's identity]], <math>1 = e^{0 \ | + | By [[Euler's identity]], <math>1 = e^{0 \times i} = \cos (0+2k\pi) + i \sin(0+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> that produce <math>24</math> unique results. | ||
+ | |||
+ | Using De Moivre's Theorem again, we have <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 to equal <math>0</math> to negate the imaginary component. This occurs whenever <math>\frac{k\pi}{2}</math> is an integer multiple of <math>\pi</math>, requiring that <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>. | |
− | + | ==Solution 3== | |
+ | From the start, recall from the Fundamental Theorem of Algebra that <math>z^{24} = 1</math> must have <math>24</math> solutions (and these must be distinct since the equation factors into <math>0 = (z-1)(z^{23} + z^{22} + z^{21}... + z + 1)</math>), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be <math>24</math>. Notice that <math>1 = z^{24} = (z^6)^4</math>, so for any solution <math>z</math>, <math>z^6</math> will be one of the 4th roots of unity (<math>1</math>, <math>i</math>, <math>-1</math>, or <math>-i</math>). Then <math>6</math> solutions <math>z</math> will satisfy <math>z^6 = 1</math>, <math>6</math> will satisfy <math>z^6 = -1</math> (and this is further justified by knowledge of the 6th roots of unity), so there must be <math>\boxed{(D) \: 12}</math> such <math>z</math>. | ||
− | + | ==Solution 4 (Quick)== | |
+ | Let <math>a\in\mathbb{R}</math> and <math>a = z^6.</math> We have | ||
+ | <cmath>a^4 = 1 \implies a = 1,-1.</cmath> | ||
+ | <math>z^6 = \pm 1</math> has 6 solutions for <math>1</math> and <math>-1</math> respectively, so <math>6+6=\boxed{(D)\ 12}.</math> <cmath> </cmath> | ||
+ | -svyn | ||
==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}} |
Latest revision as of 04:19, 2 February 2021
Problem
There are different complex numbers
such that
. For how many of these is
a real number?
Solution 1
Note that these such that
are
for integer
. So
This is real if 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
that produce
unique results.
Using De Moivre's Theorem again, we have
For to be real,
has to equal
to negate the imaginary component. This occurs whenever
is an integer multiple of
, requiring that
is even. There are exactly
even values of
on the interval
, so the answer is
.
Solution 3
From the start, recall from the Fundamental Theorem of Algebra that must have
solutions (and these must be distinct since the equation factors into
), or notice that the question is simply referring to the 24th roots of unity, of which we know there must be
. Notice that
, so for any solution
,
will be one of the 4th roots of unity (
,
,
, or
). Then
solutions
will satisfy
,
will satisfy
(and this is further justified by knowledge of the 6th roots of unity), so there must be
such
.
Solution 4 (Quick)
Let and
We have
has 6 solutions for
and
respectively, so
-svyn
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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