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2012 AIME I Problems/Problem 6

Problem 6

The complex numbers $z$ and $w$ satisfy $z^{13} = w,$ $w^{11} = z,$ and the imaginary part of $z$ is $\sin{\frac{m\pi}{n}}$, for relatively prime positive integers $m$ and $n$ with $m Find $n.$

Solutions

Solution 1

Substituting the first equation into the second, we find that $(z^{13})^{11} = z$ and thus $z^{142} = 1.$ So $z$ must be a $142$nd root of unity, and thus the imaginary part of $z$ will be $\sin{\frac{2m\pi}{142}} = \sin{\frac{m\pi}{71}}$ for some $m$ with $0 \le m < 142.$ But note that $71$ is prime and $m<71$ by the conditions of the problem, so the denominator in the argument of this value will always be $71$ and thus $n = \boxed{071.}$

Solution 2

Note that $w^{143}=w$ and similar for $z$, and they are not equal to $0$ because the question implies the imaginary part is positive. Thus $w^{142}=z^{142}=1$, so each is of the form $\sin(2 \pi k/142)$ where $k$ is a positive integer between $0$ and $141$ inclusive. This simplifies to $sin(pi*k/71)$, and $071$ is prime, so it is the only possible denominator, and thus is the answer.

 2012 AIME I (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions