Difference between revisions of "2018 AIME I Problems/Problem 6"
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==Solution 1== | ==Solution 1== | ||
− | Let <math>a=z^{120}</math>. This simplifies the problem constraint to <math>a^6-a \in \mathbb{R}</math>. This is true if <math>Im(a^6)=Im(a)</math>. Let <math>\theta</math> be the angle <math>a</math> makes with the positive x-axis. Note that there is exactly one <math>a</math> for each angle <math>0\le\theta<2\pi</math>. This must be true for <math>12</math> values of <math>a</math> (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time <math>\sin\theta=\sin{6\theta}</math>). For each of these solutions for <math>a</math>, there are necessarily <math>120</math> solutions for <math>z</math>. Thus, there are <math>12*120=1440</math> solutions for <math>z</math>, yielding an answer of <math>\boxed{440}</math>. | + | Let <math>a=z^{120}</math>. This simplifies the problem constraint to <math>a^6-a \in \mathbb{R}</math>. This is true if <math>\text{Im}(a^6)=\text{Im}(a)</math>. Let <math>\theta</math> be the angle <math>a</math> makes with the positive x-axis. Note that there is exactly one <math>a</math> for each angle <math>0\le\theta<2\pi</math>. This must be true for <math>12</math> values of <math>a</math> (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time <math>\sin\theta=\sin{6\theta}</math>). For each of these solutions for <math>a</math>, there are necessarily <math>120</math> solutions for <math>z</math>. Thus, there are <math>12*120=1440</math> solutions for <math>z</math>, yielding an answer of <math>\boxed{440}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 13:58, 16 June 2018
Contents
[hide]Problem
Let be the number of complex numbers
with the properties that
and
is a real number. Find the remainder when
is divided by
.
Solution 1
Let . This simplifies the problem constraint to
. This is true if
. Let
be the angle
makes with the positive x-axis. Note that there is exactly one
for each angle
. This must be true for
values of
(it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time
). For each of these solutions for
, there are necessarily
solutions for
. Thus, there are
solutions for
, yielding an answer of
.
Solution 2
The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to . Since
, let
, then we can write the imaginary part of
. Using the sum-to-product formula, we get
or
. The former yields
solutions, and the latter yields
solutions, giving a total of
solution, so our answer is
.
Solution 3
As mentioned in solution one, for the difference of two complex numbers to be real, their imaginary parts must be equal. We use polar form of complex numbers. Let . We have two cases to consider. Either
, or
and
are reflections across the imaginary axis.
If
, then
. Thus,
or
, giving us 600 solutions.
For the second case,
. This means
, giving us 840 solutions.
Our total count is thus
, yielding a final answer of
.
See also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.