Question

设 $N$ 为满足 $|z| = 1$ 且 $z^{6!} - z^{5!}$ 为实数的复数 $z$ 的个数。求 $N$ 除以 $1000$ 的余数。

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 $1000.$

答案

答案：440

知识点：复数三角学分类讨论

难度评级：2720

解答：

令 $z = e^{i\theta}$ ，其中 $\theta \in [0, 2\pi)$ 。那么 $z^{720} - z^{120}$ 为实数，当且仅当 $\sin 720\theta = \sin 120\theta$ ，这发生在两角模 $2\pi$ 同余或互补时：要么 $720\theta = 120\theta + 2\pi k$ ，得到 $\theta = \frac{\pi k}{300}$ ，要么 $720\theta = \pi - 120\theta + 2\pi k$ ，得到 $\theta = \frac{(2k+1)\pi}{840}$ 。

第一族在 $[0, 2\pi)$ 中有 $600$ 个值，第二族有 $840$ 个。它们不可能重合：若 $\frac{\pi k}{300} = \frac{(2j+1)\pi}{840}$ ，则 $14k = 5(2j + 1)$ ，等式左右一边为偶数、一边为奇数。

因此 $N = 600 + 840 = 1440$ ，余数为 $440$ 。

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 第 6 题

答案：440

难度评级：2720

解答：

其他年份的第 6 题