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

Revision as of 18:50, 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 $\boxed{440}$ .

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 {{AIME box|year=2018|n=I|num-b=3|num-a=5}}
 {{MAA Notice}}
-[[Category:Introductory Geometry Problems]]

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

Revision as of 18:50, 7 March 2018

Problem

Solution

See also