Difference between revisions of "2018 AIME I Problems/Problem 6"
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==Solution 6 (Official MAA)== | ==Solution 6 (Official MAA)== | ||
− | If <math>z</math> satisfies the given conditions, there is a <math>\theta \in [0,2\pi]</math> such that <math>z=e^{i\theta}</math> and <math>e^{720\theta i-120\theta i}</math> is real. This difference is real if and only if either the two numbers <math>720\theta</math> and <math>120\theta</math> represent the same angle or the two numbers represent supplementary angles. In the first case there is an integer <math>k</math> such that <math>720\theta=120\theta+2k\pi,</math> which implies that <math>\theta</math> is a multiple of <math>\tfrac{\pi}{300}.</math> In the second case there is an integer <math>k</math> such that <math>720\theta=120\theta+(2k+1)\pi,</math> which implies that <math>\theta</math> is <math>\tfrac{\pi}{840}</math> plus a multiple of <math>\tfrac{\pi}{420}.</math> In the interval <math>[0,2\pi]</math> there are <math>600</math> values of <math>\theta</math> that are multiples of <math>\tfrac{\pi}{300},</math> there are <math>840</math> values that are <math>\tfrac{\pi}{840}</math> plus a multiple of <math>\tfrac{\pi}{420},</math> and there are no values of <math>\theta</math> that satisfy both of these conditions. Therefore there | + | If <math>z</math> satisfies the given conditions, there is a <math>\theta \in [0,2\pi]</math> such that <math>z=e^{i\theta}</math> and <math>e^{720\theta i-120\theta i}</math> is real. This difference is real if and only if either the two numbers <math>720\theta</math> and <math>120\theta</math> represent the same angle or the two numbers represent supplementary angles. In the first case there is an integer <math>k</math> such that <math>720\theta=120\theta+2k\pi,</math> which implies that <math>\theta</math> is a multiple of <math>\tfrac{\pi}{300}.</math> In the second case there is an integer <math>k</math> such that <math>720\theta=-120\theta+(2k+1)\pi,</math> which implies that <math>\theta</math> is <math>\tfrac{\pi}{840}</math> plus a multiple of <math>\tfrac{\pi}{420}.</math> In the interval <math>[0,2\pi]</math> there are <math>600</math> values of <math>\theta</math> that are multiples of <math>\tfrac{\pi}{300},</math> there are <math>840</math> values that are <math>\tfrac{\pi}{840}</math> plus a multiple of <math>\tfrac{\pi}{420},</math> and there are no values of <math>\theta</math> that satisfy both of these conditions. Therefore there |
must be <math>600+840=1440</math> complex numbers satisfying the given conditions. The requested remainder is <math>440.</math> | must be <math>600+840=1440</math> complex numbers satisfying the given conditions. The requested remainder is <math>440.</math> | ||
+ | |||
==Video Solution== | ==Video Solution== | ||
Revision as of 15:00, 17 January 2022
Contents
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
. We are given
. Note that
and
. We can use this to create two types of solutions:
which reduces to . There are 7 solutions for this.
which reduces to . There are 5 solutions for this, totalling 12 values of
.
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 exponential 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. (Equalities are taken modulo
)
For the second case,
. This means
, giving us 840 solutions.
Our total count is thus
, yielding a final answer of
.
Solution 4
Because we know that
Hence
Because
is real, it is equal to its complex conjugate. Hence
Substituting the expression we that we derived earlier, we get
This leaves us with a polynomial whose leading term is
Hence our answer is
.
Solution 5
Since , let
. For
to be real, the imaginary parts of
and
must be equal, so
. We need to find all solutions for
in the interval
. This can be done by graphing
and
and finding their intersections. Since the period of
is
and the period of
is
, the common period of both graphs is
. Therefore, we only graph the functions in the domain
. We can clearly see that there are twelve points of intersection. However, since we only graphed
of the interval
, we need to multiply our answer by
. The answer is
.
Solution 6 (Official MAA)
If satisfies the given conditions, there is a
such that
and
is real. This difference is real if and only if either the two numbers
and
represent the same angle or the two numbers represent supplementary angles. In the first case there is an integer
such that
which implies that
is a multiple of
In the second case there is an integer
such that
which implies that
is
plus a multiple of
In the interval
there are
values of
that are multiples of
there are
values that are
plus a multiple of
and there are no values of
that satisfy both of these conditions. Therefore there
must be
complex numbers satisfying the given conditions. The requested remainder is
Video Solution
https://www.youtube.com/watch?v=iE8paW_ICxw
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.