24 different complex numbers

There are $24$ different complex numbers $z$ such that $z^{24}=1$. For how many of these is $z^6$ a real number?

This is one of the AMC problems from this year. I've been trying to solve it, but I couldn't and a peek at the answers (not recommended, I know) talked about Euler's theorem etc., which I haven't learnt yet. Is there an 'easier' way to solve this problem?

Let's say $w=z^6$. We know that $w^4=1$, so $w=\pm 1,\pm i$. Each of these four numbers has $6$ distinct sixth roots.

Does that help?

The $n$th complex roots of $1$ are the points on the complex unit circle with angles of $\frac{2\pi k}n, k\in \Bbb N$. This looks something like this (for $n=7$):

Now imagine this picture with $n=24$, and $n=6$. Where do they overlap?

Since $|z|=1$ for all $z$ such that $z^{24}=1,$ the only two real numbers that $z^{6}$ can possibly be are the ones with magnitude $1,$ specifically $1$ and $-1.$ Since there are $6$ solutions to $z^{6}=1$ and $6$ more to $z^{6}=-1,$ there must be $\boxed{12}$ total $z$ satisfying the conditions.

Alternatively, consider that $(z^6)^4=z^{24}=1.$ Now, if we set $z^6=w,$ then for every $w$ there are $6$ solutions in $z.$ Since the solutions for $w$ are $\pm 1, \pm i,$ half of the $24$ solutions for $z$ satisfy $w$ being a real number, so the answer is again $\boxed{12}.$