Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AIME I Problems/Problem 6"

(→‎Solution 4)
(I made this solution myself - looking back on it, it makes no sense...)
 
(23 intermediate revisions by 14 users not shown)
Line 3: Line 3:
  
 
==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>. 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>. 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==
Line 10: Line 20:
 
== 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 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.
 
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 ==
 
== 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>.
 
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 23:23, 19 January 2024

Problem

Let $N$ be the number of complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$.

Solution 1

Let $a=z^{120}$. This simplifies the problem constraint to $a^6-a \in \mathbb{R}$. This is true if $\text{Im}(a^6)=\text{Im}(a)$. Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$. We are given $\sin\theta = \sin{6\theta}$. Note that $\sin \theta = \sin (\pi - \theta)$ and $\sin \theta = \sin (\theta + 2\pi)$. We can use these facts to create two types of solutions:

\[\sin \theta = \sin ((2m + 1)\pi - \theta)\]

which implies that $(2m+1)\pi-\theta = 6\theta$ and reduces to $\frac{(2m + 1)\pi}{7} = \theta$. There are 7 solutions for this.

\[\sin \theta = \sin (2n\pi + \theta)\]

which implies that $2n\pi+\theta=6\theta$ and reduces to $\frac{2n\pi}{5} = \theta$. There are 5 solutions for this, totaling 12 values of $a$.

For each of these solutions for $a$, there are necessarily $120$ solutions for $z$. Thus, there are $12\cdot 120=1440$ solutions for $z$, yielding an answer of $\boxed{440}$.

Solution 2

The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to $0$. Since $|z|=1$, let $z=\cos \theta + i\sin \theta$, then we can write the imaginary part of $\Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0$. Using the sum-to-product formula, we get $\sin\left(720\theta\right)-\sin\left(120\theta\right)=2\cos\left(\frac{720\theta+120\theta}{2}\right)\sin\left(\frac{720\theta-120\theta}{2}\right)=2\cos\left(\frac{840\theta}{2}\right)\sin\left(\frac{600\theta}{2}\right)\implies \cos\left(\frac{840\theta}{2}\right)=0$ or $\sin\left(\frac{600\theta}{2}\right)=0$. The former yields $840$ solutions, and the latter yields $600$ solutions, giving a total of $840+600=1440$ solution, so our answer is $\boxed{440}$.

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 $z = e^{i \theta}$. We have two cases to consider. Either $z^{6!} = z^{5!}$, or $z^{6!}$ and $z^{5!}$ are reflections across the imaginary axis. If $z^{6!} = z^{5!}$, then $e^{6! \theta i} = e^{5! \theta i}$. Thus, $720 \theta = 120 \theta$ or $600\theta = 0$, giving us 600 solutions. (Equalities are taken modulo $2 \pi$) For the second case, $e^{6! \theta i} = e^{(\pi - 5!\theta)i}$. This means $840 \theta = \pi$, giving us 840 solutions. Our total count is thus $1440$, yielding a final answer of $\boxed{440}$.

Solution 4

Because $|z| = 1,$ we know that $z\overline{z} = 1^2 = 1.$ Hence $\overline{z} = \frac 1 {z}.$ Because $z^{6!}-z^{5!}$ is real, it is equal to its complex conjugate. Hence $z^{6!}-z^{5!} = \overline{z^{6!}}-\overline{z^{5!}}.$ Substituting the expression we that we derived earlier, we get $z^{720}-z^{120} = \frac 1{z^{720}} - \frac 1{z^{120}}.$ This leaves us with a polynomial whose leading term is $z^{1440}.$ Hence our answer is $\boxed{440}$.

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 $(z^{840}+1)(z^{600}-1)=0,$ so this implies that $z^{840} = -1$ OR $z^{600}=1.$ To show no $z$ satisfies both of these conditions, notice that if $w^{840} = -1$ and $w^{600}=1$ for some complex number $w$, then $w^{240} = w^{840-600} = -1,$ which implies that $w^{360} = w^{600-240}=-1,$ which implies that $w^{120}=w^{360-240}=1.$ This is a contradiction since then $w^{240}$ would also have to equal $(w^{120})^2 = 1.$ Therefore the total number of solutions is $840+600 = 1\boxed{440}.$

Solution 5

Since $|z|=1$, let $z=\cos \theta + i\sin \theta$. For $z^{6!}-z^{5!}$ to be real, the imaginary parts of $z^{6!}$ and $z^{5!}$ must be equal, so $\sin 720\theta=\sin 120\theta$. We need to find all solutions for $\theta$ in the interval $[0,2\pi)$. This can be done by graphing $y=\sin 720\theta$ and $y=\sin 120\theta$ and finding their intersections. Since the period of $y=\sin 720\theta$ is $\frac{\pi}{360}$ and the period of $y=\sin 120\theta$ is $\frac{\pi}{60}$, the common period of both graphs is $\frac{\pi}{60}$. Therefore, we only graph the functions in the domain $[0, \frac{\pi}{60})$. We can clearly see that there are twelve points of intersection. However, since we only graphed $\frac{1}{120}$ of the interval $[0,2\pi)$, we need to multiply our answer by $120$. The answer is $12 \cdot 120 = 1440 = \boxed{440} \pmod{1000}$.

Solution 6 (Official MAA)

If $z$ satisfies the given conditions, there is a $\theta \in [0,2\pi]$ such that $z=e^{i\theta}$ and $e^{720\theta i-120\theta i}$ is real. This difference is real if and only if either the two numbers $720\theta$ and $120\theta$ represent the same angle or the two numbers represent supplementary angles. In the first case there is an integer $k$ such that $720\theta=120\theta+2k\pi,$ which implies that $\theta$ is a multiple of $\tfrac{\pi}{300}.$ In the second case there is an integer $k$ such that $720\theta=-120\theta+(2k+1)\pi,$ which implies that $\theta$ is $\tfrac{\pi}{840}$ plus a multiple of $\tfrac{\pi}{420}.$ In the interval $[0,2\pi]$ there are $600$ values of $\theta$ that are multiples of $\tfrac{\pi}{300},$ there are $840$ values that are $\tfrac{\pi}{840}$ plus a multiple of $\tfrac{\pi}{420},$ and there are no values of $\theta$ that satisfy both of these conditions. Therefore there must be $600+840=1440$ complex numbers satisfying the given conditions. The requested remainder is $440.$


Solution 7

$z^{720}-z^{120}=z^{120}(z^{600}-1)$ firstly we consider that $z^{600}-1=0$ obviously there are $600$ different roots

Then, we consider that $z^{720}=a+bi, z^{120}=-a+bi$ or $z^{720}=-a-bi, z^{120}=a-bi$ which leads to $840\theta=\pi+2k\pi, k\in N$ so there are $840$ roots, in all, $840+600 \equiv 440 (mod1000)$ 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_6&oldid=211463"