10. (HUN 2) Three persons $A, B, C$, are playing the following game: A $k$ element subset of the set $\{1, \ldots, 1986\}$ is randomly chosen, with an equal probability of each choice, where $k$ is a fixed positive integer less than or equal to 1986. The winner is $A, B$ or $C$, respectively, if the sum of the chosen numbers leaves a remainder of 0, 1, or 2 when divided by 3. For what values of $k$ is this game a fair one? (A game is fair if the three outcomes are equally probable.)
10. The set $X=\{1, \ldots, 1986\}$ splits into triads $T_{1}, \ldots, T_{662}$, where $T_{j}=$ $\{3 j-2,3 j-1,3 j\}$. Let $\mathcal{F}$ be the family of all $k$-element subsets $P$ such that $\left|P \cap T_{j}\right|=1$ or 2 for some index $j$. If $j_{0}$ is the smallest such $j_{0}$, we define $P^{\prime}$ to be the $k$-element set obtained from $P$ by replacing the elements of $P \cap T_{j_{0}}$ by the ones following cyclically inside $T_{j_{0}}$. Let $s(P)$ denote the remainder modulo 3 of the sum of elements of $P$. Then $s(P), s\left(P^{\prime}\right), s\left(P^{\prime \prime}\right)$ are distinct, and $P^{\prime \prime \prime}=P$. Thus the operator ${ }^{\prime}$ gives us a bijective correspondence between the sets $X \in \mathcal{F}$ with $s(P)=0$, those with $s(P)=1$, and those with $s(P)=2$. If $3 \nmid k$ is not divisible by 3, then each $k$-element subset of $X$ belongs to $\mathcal{F}$, and the game is fair. If $3 \mid k$, then $k$-element subsets not belonging to $\mathcal{F}$ are those that are unions of several triads. Since every such subset has the sum of elements divisible by 3, it follows that player $A$ has the advantage.