2. $f(n)$ 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 $f(1789)$.
(1989 Australian Olympiad Problem)
$\begin{array}{l}\text { 2. Since } 1789=9+4 \cdot 9+4^{2} \cdot 9+4^{3} \cdot 9+4^{4} \cdot 2^{2} \text {, and } f(4 n+9)=f[f(f(n))]=4 f(n)+9 \text {, } \\ \text { therefore } f(1789)=9+4 \cdot f\left(9+4 \cdot 9+4^{2} \cdot 9+4^{3} \cdot 2^{2}\right)=9+4\left[9+4 \cdot f\left(9+4 \cdot 9+4^{2} \cdot 2^{2}\right)\right] \\ =9+4 \cdot 9+4^{2} \cdot f\left(9+4 \cdot 9+4^{2} \cdot 2^{2}\right)=9+4 \cdot 9+4^{2} \cdot 9+4^{3} \cdot f\left(9+4 \cdot 2^{2}\right) \\ =9+4 \cdot 9+4^{2} \cdot 9+4^{3} \cdot 9+4^{4} \cdot f\left(2^{2}\right)=9+4 \cdot 9+4^{2} \cdot 9+4^{3} \cdot 9+4^{4} \cdot\left(2^{3}+3\right) \\ =3581 \text {. }\end{array}$