Difference between revisions of "2018 AIME I Problems/Problem 6"
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(I made this solution myself - looking back on it, it makes no sense...) |
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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>\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>. | + | 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>. We are given <math>\sin\theta = \sin{6\theta}</math>. Note that <math>\sin \theta = \sin (\pi - \theta)</math> and <math>\sin \theta = \sin (\theta + 2\pi)</math>. We can use these facts to create two types of solutions: |
+ | |||
+ | <cmath>\sin \theta = \sin ((2m + 1)\pi - \theta)</cmath> | ||
+ | |||
+ | which implies that <math>(2m+1)\pi-\theta = 6\theta</math> and reduces to <math>\frac{(2m + 1)\pi}{7} = \theta</math>. There are 7 solutions for this. | ||
+ | |||
+ | <cmath>\sin \theta = \sin (2n\pi + \theta)</cmath> | ||
+ | |||
+ | which implies that <math>2n\pi+\theta=6\theta</math> and reduces to <math>\frac{2n\pi}{5} = \theta</math>. There are 5 solutions for this, totaling 12 values of <math>a</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\cdot 120=1440</math> solutions for <math>z</math>, yielding an answer of <math>\boxed{440}</math>. | ||
==Solution 2== | ==Solution 2== | ||
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== Solution 3 == | == 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 | + | 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 <math>z = e^{i \theta}</math>. We have two cases to consider. Either <math>z^{6!} = z^{5!}</math>, or <math>z^{6!}</math> and <math>z^{5!}</math> are reflections across the imaginary axis. |
− | If <math>z^{6!} = z^{5!}</math>, then <math>e^{6! \theta i} = e^{5! \theta i}</math>. Thus, <math>720 \theta = 120 \theta</math> or <math>600\theta = 0</math>, giving us 600 solutions. | + | If <math>z^{6!} = z^{5!}</math>, then <math>e^{6! \theta i} = e^{5! \theta i}</math>. Thus, <math>720 \theta = 120 \theta</math> or <math>600\theta = 0</math>, giving us 600 solutions. (Equalities are taken modulo <math>2 \pi</math>) |
For the second case, <math>e^{6! \theta i} = e^{(\pi - 5!\theta)i}</math>. This means <math>840 \theta = \pi </math>, giving us 840 solutions. | For the second case, <math>e^{6! \theta i} = e^{(\pi - 5!\theta)i}</math>. This means <math>840 \theta = \pi </math>, giving us 840 solutions. | ||
Our total count is thus <math>1440</math>, yielding a final answer of <math>\boxed{440}</math>. | Our total count is thus <math>1440</math>, yielding a final answer of <math>\boxed{440}</math>. | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | Because <math>|z| = 1,</math> we know that <math>z\overline{z} = 1^2 = 1.</math> Hence <math>\overline{z} = \frac 1 {z}.</math> Because <math>z^{6!}-z^{5!}</math> is real, it is equal to its complex conjugate. Hence <math>z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.</math> Substituting the expression we that we derived earlier, we get <math>z^{720}-z^{120} = \frac 1{z^{720}} - \frac 1{z^{120}}.</math> This leaves us with a polynomial whose leading term is <math>z^{1440}.</math> Hence our answer is <math>\boxed{440}</math>. | ||
+ | |||
+ | Note: This is actually not rigorous, because how to we know that all of the roots of such a polynomial are distinct? One can proceed as follows. Factoring gives us that <math>(z^{840}+1)(z^{600}-1)=0,</math> so this implies that <math>z^{840} = -1</math> OR <math>z^{600}=1.</math> To show no <math>z</math> satisfies both of these conditions, notice that if <math>w^{840} = -1</math> and <math>w^{600}=1</math> for some complex number <math>w</math>, then <math>w^{240} = w^{840-600} = -1,</math> which implies that <math>w^{360} = w^{600-240}=-1,</math> which implies that <math>w^{120}=w^{360-240}=1.</math> This is a contradiction since then <math>w^{240}</math> would also have to equal <math>(w^{120})^2 = 1.</math> Therefore the total number of solutions is <math>840+600 = 1\boxed{440}.</math> | ||
+ | |||
+ | == Solution 5 == | ||
+ | Since <math>|z|=1</math>, let <math>z=\cos \theta + i\sin \theta</math>. For <math>z^{6!}-z^{5!}</math> to be real, the imaginary parts of <math>z^{6!}</math> and <math>z^{5!}</math> must be equal, so <math>\sin 720\theta=\sin 120\theta</math>. We need to find all solutions for <math>\theta</math> in the interval <math>[0,2\pi)</math>. This can be done by graphing <math>y=\sin 720\theta</math> and <math>y=\sin 120\theta</math> and finding their intersections. Since the period of <math>y=\sin 720\theta</math> is <math>\frac{\pi}{360}</math> and the period of <math>y=\sin 120\theta</math> is <math>\frac{\pi}{60}</math>, the common period of both graphs is <math>\frac{\pi}{60}</math>. Therefore, we only graph the functions in the domain <math>[0, \frac{\pi}{60})</math>. We can clearly see that there are twelve points of intersection. However, since we only graphed <math>\frac{1}{120}</math> of the interval <math>[0,2\pi)</math>, we need to multiply our answer by <math>120</math>. The answer is <math>12 \cdot 120 = 1440 = \boxed{440} \pmod{1000}</math>. | ||
+ | |||
+ | ==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 | ||
+ | must be <math>600+840=1440</math> complex numbers satisfying the given conditions. The requested remainder is <math>440.</math> | ||
+ | |||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | <math>z^{720}-z^{120}=z^{120}(z^{600}-1)</math> firstly we consider that <math>z^{600}-1=0</math> obviously there are <math>600</math> different roots | ||
+ | |||
+ | Then, we consider that <math>z^{720}=a+bi, z^{120}=-a+bi</math> or <math>z^{720}=-a-bi, z^{120}=a-bi</math> which leads to <math>840\theta=\pi+2k\pi, k\in N</math> so there are <math>840</math> roots, in all, <math>840+600 \equiv 440 (mod1000)</math> we are done | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=iE8paW_ICxw | ||
== See also == | == See also == | ||
{{AIME box|year=2018|n=I|num-b=5|num-a=7}} | {{AIME box|year=2018|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:23, 20 January 2024
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 these facts to create two types of solutions:
which implies that and reduces to
. There are 7 solutions for this.
which implies that and reduces to
. There are 5 solutions for this, totaling 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
.
Note: This is actually not rigorous, because how to we know that all of the roots of such a polynomial are distinct? One can proceed as follows. Factoring gives us that so this implies that
OR
To show no
satisfies both of these conditions, notice that if
and
for some complex number
, then
which implies that
which implies that
This is a contradiction since then
would also have to equal
Therefore the total number of solutions 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
Solution 7
firstly we consider that
obviously there are
different roots
Then, we consider that or
which leads to
so there are
roots, in all,
we are done
~bluesoul
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.