1. The problem asks for the number of generators of a cyclic group of order 800. 2. A cyclic group of order $n$ has exactly $\varphi(n)$ generators, where $\varphi$ is Euler's totient function. 3. We need to calculate $\varphi(800)$. 4. First, factorize 800 into primes: $$800 = 2^5 \times 5^2$$ 5. Euler's totient function for $n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$ is $$\varphi(n) = n \left(1 - \frac{1}{p_1}\right) \left(1 - \frac{1}{p_2}\right) \cdots \left(1 - \frac{1}{p_k}\right)$$ 6. Applying this to $800 = 2^5 \times 5^2$, we get: $$\varphi(800) = 800 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{5}\right)$$ 7. Simplify step-by-step: $$\varphi(800) = 800 \times \frac{1}{2} \times \frac{4}{5}$$ 8. Calculate intermediate values: $$800 \times \frac{1}{2} = \cancel{800} \times \frac{1}{\cancel{2}} = 400$$ $$400 \times \frac{4}{5} = \cancel{400} \times \frac{4}{\cancel{5}} = 80 \times 4 = 320$$ 9. Therefore, the number of generators is $\boxed{320}$. 10. The correct answer is option c. 320.