2018 AIME I Problem 6

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

Write

z = e^{i\theta};

the difference is real exactly when

\sin 720\theta = \sin 120\theta

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

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.

Problem 6 in other years