(USAMO, 1992). Let $a_{n}$ be the number written in base 10 with $2^{n}$ digits "9", for $n \geqslant 0$. Let $b_{n}=a_{0} a_{1} \cdots a_{n}$ be the product of the first $a_{k}$. What is the sum of the digits of $b_{n}$?
We denote $S_{n}$ as the sum of the digits of $b_{n}$. We have $S_{0}=9$. We have $b_{n+1}=a_{n+1} b_{n}=$ $10^{2^{n+1}} b_{n}-b_{n}$. Knowing that $a_{n}$ has $2^{n}$ digits, $b_{n}$ has at most $2^{n+1}-1$ digits (note that the number of digits of a product is at most the sum of the number of digits of the factors).

The decomposition

$$
b_{n+1}=10^{2^{n+1}}\left(b_{n}-1\right)+\left(10^{2^{n+1}}-b_{n}\right)
$$

allows us to find the $2^{n+1}$ digits of the first half and the second. Thus, we find $S_{n+1}=$ $\left(S_{n}-1\right)+\left(9 \times 2^{n+1}-S_{n}+1\right)=9 \times 2^{n+1}$.