$6 \cdot 3 f(n)$ is defined on the set of positive integers, and
(1) for any positive integer $n, f(f(n))=4 n+9$;
(2) for any non-negative integer $k, f\left(2^{k}\right)=2^{k+1}+3$. Determine $f(1789)$.
[Solution] $1789=9+4 \times 9+4^{2} \times 9+4^{3} \times 9+4^{4} \times 2^{2}$,
and $\quad f(4 n+9)=f(f(f(n)))=4 f(n)+9$
Therefore, $f(1789)=9+4 f\left(9+4 \times 9+4^{2} \times 9+4^{3} \times 2^{2}\right)$
$=9+4 \times 9+4^{2} f\left(9+4 \times 9+4^{2} \times 2^{2}\right)$
$=9+4 \times 9+4^{2} \times 9+4^{3} f\left(9+4 \times 2^{2}\right)$
$=9+4 \times 9+4^{2} \times 9+4^{3} \times 9+4^{4} f\left(2^{2}\right)$
$=9+4 \times 9+4^{2} \times 9+4^{3} \times 9+4^{4} \times\left(2^{3}+3\right)$
$=1789+1792$
$=3581$.