CONO-SUR 1995 Problem 6

Let $n$ be a natural number and $f(n) = 2n - 1995 \lfloor \frac{n}{1000} \rfloor$ ( $\lfloor$ $\rfloor$ denotes the floor function).

1. Show that if for some integer $r$ : $f(f(f...f(n)...))=1995$ (where the function $f$ is applied $r$ times), then $n$ is multiple of $1995$ .

2. Show that if $n$ is multiple of 1995, then there exists r such that: $f(f(f...f(n)...))=1995$ (where the function $f$ is applied $r$ times). Determine $r$ if $n=1995.500=997500$