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

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==Solution==
 
==Solution==
 
Let <math>a=z^{120}</math>. This simplifies the problem constraint to <math>a^6-a \in \mathbb{R}</math>. This is true iff <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 iff <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>.
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== See also ==
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{{AIME box|year=2018|n=I|num-b=3|num-a=5}}
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{{MAA Notice}}
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[[Category:Introductory Geometry Problems]]

Revision as of 19:49, 7 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

Let $a=z^{120}$. This simplifies the problem constraint to $a^6-a \in \mathbb{R}$. This is true iff $Im(a^6)=Im(a)$. 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 $\boxed{440}$.

See also

2018 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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