$A, B$ and $C$ played cards. First, $A$ won; $B$ and $C$ each lost a third of their money. Second, $B$ won; this time $A$ and $C$ each lost a third of their money. Finally, $C$ won and this time $A$ and $B$ each lost a third of their money. At the end of the game, $A$ had a loss of $2 \mathrm{~K}$; $C$ won twice as much as the amount of money he had at the beginning of the game plus 8 crowns. How much money did each of them have, if the total amount of their money at the beginning was less than 1000 crowns.
At the end of the third game

$$
\begin{array}{r}
A \text { had } \frac{12 x+4 y+4 z}{27} \\
B \text { had } \frac{114 y+6 x+6 z}{27} \\
C \text { had } \frac{17 z+9 x+9 y}{27} \text { crowns. }
\end{array}
$$

Therefore, according to the problem:

$$
x-\frac{12 x+4 y+4 z}{27}=2
$$

and

$$
\frac{9 x+9 y+17 z}{27}-z=2 z+8
$$

Rearranging these equations, we get

$$
15 x-4 y-4 z=54
$$

and

$$
9 x+9 y-64 z=216
$$

Solving this system of indeterminate equations, we obtain

$$
\begin{aligned}
& x=292 a+54 \\
& y=924 a+162 \\
& x=171 a+27
\end{aligned}
$$

But according to the problem, $x+y+z<1000 \text{ s}$, so it is necessary that $a=0$. Therefore, $x=54, y=162, z=27$.

Thus, $A$ had 54, $B$ had 162, and $C$ had 27 crowns.

(Jánosy Gyula, Budapest.)

The problem was also solved by: Bartók I., Biró A., Brambring V., Deutsch I., Enyedi B., Haar A., Kertész G., Kiss J., Kürti I., Liebner A., Pám M., Pivnyik I., Riesz K., Schlesinger O., Schöffer I., Schwemmer I.