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 2)
(→‎Solution 3)
Line 9: Line 9:
  
 
== 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 polar form of complex numbers. Let <math>z = e^{i \theta}</math>. Now, on the complex plane, this means there are two cases to consider: Either <math>z^{6!} = z^{5!}</math> (they're equal), or the two complex numbers are reflections across the Imaginary axis (the imaginary parts are equal, real parts are negative each other).  
+
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 <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!} 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 \equiv 120 \theta \mod 2\pi<math> or </math>600\theta \equiv 0 \mod 2\pi<math>, giving us 600 solutions.
−
If the <math>z^{6!} = z^{5!}</math>, then <math>e^{6! \theta i} = e^{5! \theta i}</math>. Thus, <math>720 \theta \equiv 120 \theta \mod 2\pi</math>.
+
For the second case, </math>e^{6! \theta i} = e^{(\pi - 5!\theta)i}<math>. This means </math>840 \theta \equiv \pi mod 2\pi<math>, giving us 840 solutions.
 +
Our total count is thus </math>\boxed{1440}$.
  
 
== 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 00:54, 20 April 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 $\boxed{440}$.

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 $z = e^{i \theta}$. 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,$720 \theta \equiv 120 \theta \mod 2\pi$or$600\theta \equiv 0 \mod 2\pi$, giving us 600 solutions. For the second case,$e^{6! \theta i} = e^{(\pi - 5!\theta)i}$. This means$840 \theta \equiv \pi mod 2\pi$, giving us 840 solutions. Our total count is thus$\boxed{1440}$.

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_6&oldid=94039"