$C$ and $D$ played based on the following agreement: 1) the loser of a won game gives the winner as many seed beads as the winner currently has, 2) the value of a game won after a draw is twice the number of beads of the game played two games ago. The winners of the consecutive games were $C, C, D$, then the game was a draw: $X$, followed by $C, X, D, C$, $X, X, D$. After the last game, $C$ refused to pay, saying that the value of the second-to-last game was 0. Therefore, $D$ did not play further and determined that they had the same number of beads as they had before the game. - How many beads did each of them have, if the smallest prize that occurred was 20 seed beads?
At the beginning of the game, $C$ had $x$ marbles and $D$ had $y$ marbles. During the subsequent games, the winnings and the number of marbles each player had evolved as follows:

| row-  number winner |  | winnings |  | C has |  | winnings |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1. | $C$ | $x$ | $(40)$ | $2 x$ | (80) | $y-x$ | $(126+2 t)$ | $6 k$ |
| 2. | $C$ | $2 x$ | $(80)$ | $4 x$ | $(160)$ | $y-3 x$ | $(46+2 t)$ | $12 k$ |
| 3. | $D$ | $y-3 x$ | $(46+2 t)$ | $7 x-y$ | $(113+t)$ | $2 y-6 x$ | $(93+t)$ | $7 k$ |
| 4. | $X$ | - | - | $7 x-y$ | $(113+t)$ | $2 y-6 x$ | $(93+t)$ | - |
| 5. | $C$ | $2 y-6 x$ | $(93+t)$ | $x+y$ | $(206+2 t)$ | 0 | $(0)$ | $14 k$ |
| 6. | $X$ | - | - | $x+y$ | $(206+2 t)$ | 0 | $(0)$ | - |
| 7. | $D$ | $4 y-12 x$ | $(186+2 t)$ | $13 x-3 y$ | $(20)$ | $4 y-12 x$ | $(186+2 t)$ | $28 k$ |
| 8. | C | $13 x-3 y$ | (20) | $26 x-6 y$ | $(40)$ | $7 y-25 x$ | $(166+2 t)$ | $3 k$ |

In the last three games, the number of marbles each player had did not change. Therefore,

$$
7 y-25 x=y, \quad 6 y=25 x
$$

In other words,

$$
x: y=6: 25, \quad \text { i.e. } \quad x=6 k, \quad y=25 k,
$$

where $k$ is a positive number. Thus, expressing the winnings in terms of $k$, we can select the smallest winnings that occurred (see the last column of the table). From $3 k=20$,

$$
x=6 k=40, \quad y=25 k=166 \frac{2}{3}
$$

The table includes the trial data in parentheses, where $t$ stands for the number $1 / 3$.

Katalin Vesztergombi (Budapest, Fazekas M. High School II. Class)

Remarks. 1. Some found $C$'s stance unfair. However, a game is determined by the rules, and the described rule for handling a series of draws is unambiguous—although it may seem surprising to some—thus, $C$'s only correct stance was the one stated. As for fairness, it is no more or less fair than doubling the stake after a draw (even if some might expect a further doubling after two draws, analogous to "doubling" and "redoubling" in bridge; the fact is, the agreement was not this).

2. Many were puzzled by the appearance of fractional marbles in the solution. Some, suspecting a printing error, sought another solution with integer values, while others deemed the problem flawed, stating, "... because there are no fractional marbles" - as two solvers wrote. In fact, a marble can naturally break, but it is also conceivable that a chipped marble was considered two-thirds of a marble by the players. Therefore, the non-integer solution is conceivable in several ways.