1. Determine the smallest natural number $n$ such that the sums of the digits of the numbers $n$ and $n+1$ are divisible by 1985.
Solution. A Baraniot number is of the form

$$
n=c 99 \ldots 9899 \ldots 9
$$

where $l$ is the smallest natural number for which $1985|9l-1$, and $k$ is the smallest natural number such that for some $c \in\{1,2, \ldots, 9\}$, $1985=9(k+1)+c$. From the condition

$$
9 l-1=1985 s=9 \cdot 220 s+5 s
$$

we obtain that $l=1544$ for $s=7$. Furthermore, it is easily obtained that $k=219, c=5$.