1. **Problem statement:** Find how many elements of the group $\mathbb{Z}_{24}$ have order 6. 2. **Recall the formula:** The order of an element $k$ in $\mathbb{Z}_n$ is given by $$\text{order}(k) = \frac{n}{\gcd(n,k)}$$ where $\gcd$ is the greatest common divisor. 3. **Apply the formula:** We want elements $k$ such that $$\frac{24}{\gcd(24,k)} = 6$$ which implies $$\gcd(24,k) = \frac{24}{6} = 4$$ 4. **Find all $k$ with $1 \leq k \leq 24$ such that $\gcd(24,k) = 4$.** 5. **Calculate:** The divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. We want $k$ where $\gcd(24,k) = 4$. 6. **Check candidates:** - $\gcd(24,4) = 4$ - $\gcd(24,8) = 8$ (no) - $\gcd(24,12) = 12$ (no) - $\gcd(24,16) = 8$ (no) - $\gcd(24,20) = 4$ 7. **List all $k$ with $\gcd(24,k) = 4$: $k = 4, 20$.** 8. **Count:** There are 2 such elements. **Final answer:** 2 elements of $\mathbb{Z}_{24}$ have order 6.