1. ## f(84)

let f function from z to z(integers)

and

f(n)=n - 3 at n > 999

f(n)= f(f(n+5) at n < 1000

find
f(84)

2. Here is a start:
$\displaystyle f(997) = f\left( {f(1002)} \right) = f(999) = f\left( {f(1004)} \right) = f(1001) = 998.$

$\displaystyle f(998) = f\left( {f(1003)} \right) = f(1000) = 997.$

One more:
$\displaystyle f(995) = f\left( {f(1000)} \right) = f(997) = 998.$

3. Note that,
$\displaystyle f(n)$ for $\displaystyle n\leq 999$ is determined by the parity of $\displaystyle n$.
Meaning $\displaystyle f(n)=998$ if $\displaystyle n$ is odd. And $\displaystyle f(n)=997$ if $\displaystyle n$ is even.

We can show this by Strong Induction.
First,
$\displaystyle f(999)=f(f(1004))=f(1001)=998$
$\displaystyle f(998)=f(f(1003))=f(1000)=997$
$\displaystyle f(997)=f(f(1002))=f(999)=998$
$\displaystyle f(996)=f(f(1001))=f(998)=997$

Now, if $\displaystyle n\leq 995$ and odd we have,
$\displaystyle f(n)=f(f(n+5))$
But by induction and the fact $\displaystyle n+5$ is even
$\displaystyle f(n)=f(997)=998$

And, if $\displaystyle n\leq 995$ and even we have,
$\displaystyle f(n)=f(f(n+5))$
But by induction and the fact $\displaystyle n+5$ is odd
$\displaystyle f(n)=f(998)=997$

The above gives us an explicit formula for f(n).