2. Function $f$ is defined on the set of integers, satisfying
$$
f(n)=\left\{\begin{array}{ll}
n-3 & n \geqslant 1000 \\
f[f(n+5)] & n<1000
\end{array},\right.
$$

Find $f(84)$.
2. Let $f_{1}(n)=f(n), f_{m+1}(n)=f_{m}(f(n)) \quad m=1,2, \cdots$, then $f(84)=f_{2}(84+5)=\cdots=f_{184}(84 \times 183 \times 5)=f_{184}(999)$.
For natural number $n$, we have
$$
\begin{array}{c}
f_{n+2}(999)=f_{n+3}(1004)=\cdots=f_{n}(997)=f_{n+1}(1002)=f_{n}(999), \\
\text { so } f(84)=f_{184}(999)=f_{182}(999)=\cdots=f_{2}(999)=f_{3}(1004) \\
=f_{2}(1001)=f(998)=f_{2}(1003)=f(1000)=997 .
\end{array}
$$