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

(Solution 3)
(14 intermediate revisions by 6 users not shown)
Line 10: Line 10:
 
== 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 ==
 +
 +
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>.
 +
 +
== 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} (mod 1000)</math>.
  
== Solution 4 ==
+
==Video Solution==
  
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>.
+
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}}

Revision as of 21:34, 23 July 2020

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 $0\le\theta<2\pi$. 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 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 $\boxed{440}$.

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 $12 \cdot 120 = 1440 = \boxed{440} (mod 1000)$.

Video Solution

https://www.youtube.com/watch?v=iE8paW_ICxw

See also

2018 AIME I (ProblemsAnswer KeyResources)
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