Question 99
The function $f$ is defined for non-negative integers and satisfies the condition
$$
f(n)= \begin{cases}f(f(n+11)), & \text{ if } n \leq 1999 \\ n-5, & \text{ if } n\gt 1999\end{cases}
$$
Find all solutions of the equation $f(n)=1999$.
$$
f(n)= \begin{cases}f(f(n+11)), & \text{ if } n \leq 1999 \\ n-5, & \text{ if } n\gt 1999\end{cases}
$$
Find all solutions of the equation $f(n)=1999$.