2018 AIME I 第 6 题

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

小提示：

写成

z = e^{i\theta}

；差为实数恰好等价于

\sin 720\theta = \sin 120\theta

Write

z = e^{i\theta};

the difference is real exactly when

\sin 720\theta = \sin 120\theta

大提示：

正弦相等意味着两角模

2\pi

同余或互补；分别数出

\theta \in [0, 2\pi)

的两类解，并检查两类不重合

Equal sines mean the angles are equal or supplementary modulo

2\pi;

count

\theta \in [0, 2\pi)

of each type and check the two families do not overlap

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

小提示：

大提示：

解答：

其他年份的第 6 题