Question

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 $1000.$

Answer

Answer: 440

Concepts:complex number trigonometry casework

Difficulty rating: 2720

Solution:

Write $z = e^{i\theta}$ with $\theta \in [0, 2\pi).$ Then $z^{720} - z^{120}$ is real exactly when $\sin 720\theta = \sin 120\theta,$ which happens when the angles are equal or supplementary modulo $2\pi:$ either $720\theta = 120\theta + 2\pi k,$ giving $\theta = \frac{\pi k}{300},$ or $720\theta = \pi - 120\theta + 2\pi k,$ giving $\theta = \frac{(2k+1)\pi}{840}.$

The first family has $600$ values in $[0, 2\pi)$ and the second has $840.$ They cannot coincide: $\frac{\pi k}{300} = \frac{(2j+1)\pi}{840}$ would give $14k = 5(2j + 1),$ equating an even number with an odd one.

Hence $N = 600 + 840 = 1440,$ and the remainder is $440.$

2018 AIME I Problem 6

Answer: 440

Difficulty rating: 2720

Solution:

Problem 6 in Other Years