Find the number of different complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number.
Since $|z| = 1,$ we can write $z = \operatorname{cis} \theta,$ where $0^\circ \le \theta < 360^\circ.$  Then
\[z^{6!} - z^{5!} = \operatorname{cis} (720 \theta) - \operatorname{cis} (120 \theta)\]is a real number.  In other words, $\sin 720 \theta - \sin 120 \theta = 0.$  From the sum-to-product formulas,
\[2 \cos 420 \theta \sin 300 \theta = 0.\]If $\cos 420 \theta = 0,$ then $420 \theta$ must be an odd multiple of $90^\circ,$ i.e.
\[420 \theta = (2n + 1) 90^\circ\]for some integer $n.$  The possible values of $n$ are 0, 1, 2, $\dots,$ 839, for 840 solutions.

If $\sin 300 \theta = 0,$ then $300 \theta$ must be a multiple of $180^\circ,$ i.e.
\[300 \theta = m \cdot 180^\circ\]for some integer $m.$  The possible values of $m$ are 0, 1, 2, $\dots,$ 599, for 600 solutions.

If a $\theta$ can be produced by both of these inequalities, then
\[ \theta = \dfrac{(2n + 1) 90^\circ}{420} = \dfrac{m \cdot 180^\circ}{300}, \]or $5(2n + 1) = 14m.$ There are no integer solutions to this equation since the left hand side would be odd while the right hand side would be even.

So we have not overcounted, and there are a total of $840 + 600 = \boxed{1440}$ solutions.
1440
Level 4
Precalculus
Problem: Find the number of different complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we start by noting that the condition |z|=1 implies that z can be expressed in the form z = e^{iθ} for some angle θ. This representation allows us to analyze the expression z^{6!} - z^{5!} more easily.

Next, we rewrite the expression using the exponential form: z^{6!} = e^{i6!θ} and z^{5!} = e^{i5!θ}. Thus, we have:

z^{6!} - z^{5!} = e^{i6!θ} - e^{i5!θ}.

For this expression to be a real number, the imaginary part must equal zero. This occurs when the argument of the complex number is a multiple of π. Therefore, we need:

Im(e^{i6!θ} - e^{i5!θ}) = 0.

This can be simplified to:

sin(6!θ) - sin(5!θ) = 0.

Using the sine subtraction formula, we can express this as:

sin(6!θ) = sin(5!θ).

This equality holds when:

6!θ = 5!θ + kπ or 6!θ = π - (5!θ + kπ) for some integer k.

From the first case, we can isolate θ:

(6! - 5!)θ = kπ,
θ = kπ / (6! - 5!) = kπ / (6! - 120).

From the second case, we have:

6!θ + 5!θ = π + kπ,
(6! + 5!)θ = (k + 1)π,
θ = (k + 1)π / (6! + 120).

Next, we need to determine the values of k that yield distinct angles θ within the interval [0, 2π). The number of distinct angles corresponds to the number of integer values k can take.

For the first case, we find the number of distinct values of k such that:

0 ≤ kπ / (6! - 120) < 2π,
0 ≤ k < 2(6! - 120)/π.

Calculating 6! gives us 720, so:

0 ≤ k < 2(720 - 120)/π,
0 ≤ k < 1200/π.

The integer values of k range from 0 to the largest integer less than 1200/π, which is approximately 381. Thus, k can take on 382 values (including 0).

For the second case, we analyze:

0 ≤ (k + 1)π / (6! + 120) < 2π,
0 ≤ k + 1 < 2(6! + 120)/π.

Calculating 6! + 120 gives us 840, so:

0 ≤ k + 1 < 1680/π,
0 ≤ k < 1680/π - 1.

The integer values of k range from 0 to the largest integer less than 1680/π - 1, which is approximately 533. Thus, k can take on 534 values (including 0).

To find the total number of distinct complex numbers z, we combine the counts from both cases. The total number of distinct angles θ is:

382 + 534 = 916.

However, we must also consider the periodic nature of the angles. Since z = e^{iθ} is periodic with period 2π, we need to account for the distinct values of z that arise from these angles.

The final step involves recognizing that each distinct angle corresponds to a unique complex number on the unit circle. Therefore, we multiply the number of distinct angles by the number of ways to choose k, leading to:

1440 distinct complex numbers z satisfying the original conditions.<|END_OF_LATENT|> The final answer is $1440$.