2023 AMC 10A Problems/Problem 19

Let $f(x)$ be the number of trailing zeroes of $x!$ . Let $g(x)=5x+1$ . Find the sum of the digits of $f(1)+f(g(1))+f(g^2(1))+...+f(g^{10}(1))$ given that $5^{11}=48828125$ . $\newline$ $\textbf{(A) } 10 \ \ \ \ \ \ \textbf{(B) } 13 \ \ \ \ \ \ \textbf{(C) } 16\ \ \ \ \ \ \textbf{(D) } 19\ \ \ \ \ \ \textbf{(E) } 22$