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"

m (→‎Solution)
(→‎Solution)
Line 2: Line 2:
 
Let <math>N</math> be the number of complex numbers <math>z</math> with the properties that <math>|z|=1</math> and <math>z^{6!}-z^{5!}</math> is a real number. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
Let <math>N</math> be the number of complex numbers <math>z</math> with the properties that <math>|z|=1</math> and <math>z^{6!}-z^{5!}</math> is a real number. Find the remainder when <math>N</math> is divided by <math>1000</math>.
  
−
==Solution==
+
==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>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>.
 +
 +
==Solution 2==
 +
The constraint mentioned in the problem is equivalent to the requirement that the imaginary part is equal to <math>0</math>. Since <math>|z|=1</math>, let <math>z=\cos \theta + i\sin \theta</math>, then we can write the imaginary part of <math> \Im(z^{6!}-z^{5!})=\Im(z^{720}-z^{120})=\sin\left(720\theta\right)-\sin\left(120\theta\right)=0</math>. Using the sum-to-product formula, we get <math>\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</math> or <math>\sin\left(\frac{600\theta}{2}\right)=0</math>. The former yields <math>840</math> solutions, and the latter yields <math>600</math> solutions, giving a total of <math>840+600=1440</math> solution, so our answer is <math>\boxed{440}</math>.
  
 
== 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}}

Revision as of 18:22, 10 March 2018

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 $Im(a^6)=Im(a)$. Let $\theta$ be the angle $a$ makes with the positive x-axis. Note that there is exactly one $a$ for each angle $0\le\theta<2\pi$. This must be true for $12$ values of $a$ (it may help to picture the reference angle making one orbit from and to the positive x-axis; note every time $\sin\theta=\sin{6\theta}$). For each of these solutions for $a$, there are necessarily $120$ solutions for $z$. Thus, there are $12*120=1440$ 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 $\boxed{440}$.

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=93127"