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

$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }6\qquad\textbf{(D) }12\qquad\textbf{(E) }24$
1. We start with the given equation \( z^{24} = 1 \). The 24th roots of unity are the solutions to this equation. These roots can be expressed using Euler's formula as:
   \[
   z = e^{2k\pi i / 24} = \cos\left(\frac{2k\pi}{24}\right) + i \sin\left(\frac{2k\pi}{24}\right)
   \]
   where \( k \) is an integer ranging from 0 to 23, giving us 24 unique complex numbers.

2. We need to determine for how many of these roots \( z \), the expression \( z^6 \) is a real number. Using De Moivre's Theorem, we have:
   \[
   z^6 = \left(e^{2k\pi i / 24}\right)^6 = e^{2k\pi i / 4} = \cos\left(\frac{2k\pi}{4}\right) + i \sin\left(\frac{2k\pi}{4}\right)
   \]

3. For \( z^6 \) to be a real number, the imaginary part must be zero. This requires:
   \[
   \sin\left(\frac{2k\pi}{4}\right) = 0
   \]
   The sine function is zero when its argument is an integer multiple of \( \pi \). Therefore, we need:
   \[
   \frac{2k\pi}{4} = m\pi \quad \text{for some integer } m
   \]
   Simplifying, we get:
   \[
   \frac{k\pi}{2} = m\pi \implies k = 2m
   \]
   Thus, \( k \) must be an even integer.

4. We now count the even integers \( k \) in the range \( 0 \leq k < 24 \). The even integers in this range are:
   \[
   0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
   \]
   There are exactly 12 even integers in this range.

5. Therefore, there are 12 values of \( k \) for which \( z^6 \) is a real number.

\(\boxed{12}\)