1. Determine the set $S$ with the smallest number of elements for which the following conditions are satisfied:
a) $S \subset\{0,1,2,3, \ldots\}$,

b) $1981 \in S$,

c) If $x, y, z \in S$, then the remainder of the division of $x+y+z$ by 1982 also belongs to $S$.
Solution. We will prove that all odd numbers $1, 3, \ldots, 1981$ belong to the set $S$. The latter follows from b) and c) based on the following congruences:

$$
\begin{aligned}
& 3 \cdot 1981=1979 \text { (mod1982), } \\
& 2 \cdot 1981+1979 \equiv 1977(\bmod 1982) \\
& 2 \cdot 1981+1977 \equiv 1975(\bmod 1982) \\
& 2 \cdot 1981+3 \equiv 1(\bmod 1982)
\end{aligned}
$$

Since the set $S=\{1,3, \ldots, 1981\}$ satisfies all conditions of the problem, it is also the set with the smallest number of elements that has the given properties.