### Instruction: Find the number of different complex numbers $z$ with the properties that $|z|=1$ and $z^{6!}-z^{5!}$ is a real number.### Response: This problem involves complex numbers, their absolute values, and their powers.
I know that $|z|=1$ means that $z$ lies on the unit circle in the complex plane, and that the argument of $z$ is the angle that $z$ makes with the positive real axis.
I also know that raising a complex number to a power multiplies its argument by that power, and that a complex number is real if and only if its argument is a multiple of $\pi$.
So, I need to find the values of $z$ such that the difference of their arguments when raised to the powers $6!$ and $5!$ is a multiple of $\pi$.
I can write $z$ in polar form as $z=e^{i\theta}$, where $\theta$ is the argument of $z$.
Then, $z^{6!}=e^{i\cdot 6!\theta}$ and $z^{5!}=e^{i\cdot 5!\theta}$.
The difference of their arguments is $6!\theta-5!\theta=(6!-5!)\theta=720\theta-120\theta=600\theta$.
I want this to be a multiple of $\pi$, so I can write $600\theta=k\pi$, where $k$ is an integer.
Solving for $\theta$, I get $\theta=\frac{k\pi}{600}$.
But, I also know that $\theta$ is the argument of $z$, which must lie between $-\pi$ and $\pi$.
So, I need to find the values of $k$ that satisfy $-\pi\leq \frac{k\pi}{600}\leq \pi$.
Dividing by $\pi$, I get $-1\leq \frac{k}{600}\leq 1$.
Multiplying by $600$, I get $-600\leq k\leq 600$.
So, $k$ can take any integer value between $-600$ and $600$, inclusive.
That means there are $1201$ possible values of $k$, and each one gives a different value of $z$.
Therefore, the number of different complex numbers $z$ that satisfy the given conditions is $1201$.