Three children: $A, B, C$ are playing a skill game, and since all three are ball collectors, they agree that the current loser will triple the ball count of the other two players (i.e., give each of them twice as many balls as they currently have). After three rounds, in which $A, B, C$ lost in turn, each player has an equal number of balls. How many balls did each originally have, if after the first round, $C$ had 54 more balls than $A$?
I. solution: Originally, there were $x, y$, and $z$ balls at $A-, B-, C$ respectively.

In the I. game, $A$ lost and the number of balls changed as follows: $x-2, y-2, z, 3y, 3z$.

In the II. game, $B$ lost, and after that, the situation was: $A$ had $3x-6, B$ had $3y-(2x-4y-4z+6z)=7y-2x-2z, C$ had $9z$.

In the III. game, $C$ lost and after that, $A$ had $9x-18y-18z, B$ had $21y-6x-6z$, and $C$ had $9z-(6x-12y-12z+14y-4x-4z)=25z-2x-2y$ balls.

According to the problem,

$$
\begin{aligned}
& 9x-18y-18z=21y-6x-6z \\
& 9x-18y-18z=25z-2x-2y
\end{aligned}
$$

and finally

$$
x-2y-2z=3z-54
$$

Combining and rearranging,

$$
\begin{aligned}
15x-39y-12z & =0 \\
11x-16y-43z & =0 \\
x-2y-5z & =-54
\end{aligned}
$$

Subtracting 15 times (6) from (4) and 11 times (6) from (5),

$$
\begin{aligned}
-9y+63z & =810 \\
6y+12z & =594
\end{aligned}
$$

Simplifying,

$$
\begin{array}{r}
-y+7z=90 \\
y+2z=99
\end{array}
$$

Adding (7') and (8'),

$$
9z=189, \quad \text { from which } \quad z=21
$$

and from (8'),

$$
y=99-2 \cdot 21=57
$$

and from (6),

$$
x=-54+2 \cdot 57+5 \cdot 21=-54+114+105=165
$$

László Mercz (Pannonhalma, Bencés g. I. o. t.)

II. solution: Reasoning in reverse chronological order with one unknown leads to a simpler solution.

After the III. game, each player had $x$ balls.

Before the III. game, the distribution of balls among $A, B, C$ was: $\frac{x}{3}, \frac{x}{3}, \frac{7x}{3}$.

Before the II. game (which $B$ lost), $A$ had $\frac{x}{9}, C$ had $\frac{7x}{9}$, and thus $B$ had $\frac{x}{3}+\frac{2x}{9}+\frac{14x}{9}=\frac{19x}{9}$.

Originally (before the I. game, which $A$ lost), $B$ had $\frac{19x}{27}, C$ had $\frac{7x}{27}$, and thus $A$ had $\frac{x}{9}+\frac{38x}{27}+\frac{14x}{27}=\frac{55x}{27}$. According to the problem,

$$
\frac{x}{9}=\frac{7x}{9}-54
$$

from which

$$
x=81
$$

Originally, therefore,

$A$ had $\frac{55 \cdot 81}{27}=165, B$ had $\frac{19 \cdot 81}{27}=57, C$ had $\frac{7 \cdot 81}{27}=21$ balls.