Three friends play a game of marbles, agreeing that the loser of each round will double the number of marbles the other players have, meaning he gives the other two a number of marbles such that they end up with twice what they had at the beginning of the round. The first player lost the first round, the second player the second, the third player the third, and they all ended up with 64 marbles each. How many marbles did each of the three friends start with?
Solution 1: Let $x, y$, and $z$ be the number of marbles each player had at the beginning of the game. We have

|  | $1^{\mathbf{o}}$ | $2^{\underline{o}}$ | $3^{\underline{o}}$ |
| :---: | :---: | :---: | :---: |
| Start | $x$ | $y$ | $z$ |
| $1^{\underline{a}}$ round | $x-y-z$ | $2 y$ | $2 z$ |
| $2^{\underline{a}}$ round | $2(x-y-z)$ | $2 y-2 z-(x-y-z)$ | $4 z$ |
| $3^{\underline{a}}$ round | $4(x-y-z)$ | $2(3 y-x-z)$ | $4 z-2(x-y-z)-(3 y-x-z)$ |

Since each player ended the game with 64 marbles, it follows that

$$
\left\{\begin{aligned}
4(x-y-z) & =64 \\
2(3 y-x-z) & =64 \\
4 z-2(x-y-z)-(3 y-x-z) & =64
\end{aligned}\right.
$$

which simplifies to

$$
\left\{\begin{aligned}
x-y-z & =16 \\
-x+3 y-z & =32 \\
-x-y+7 z & =64
\end{aligned}\right.
$$

To solve the system, we add the first equation to the second and the first to the third, obtaining

$$
\left\{\begin{aligned}
y-z & =24 \\
-y+3 z & =40
\end{aligned}\right.
$$

From this, we get $z=32$ and $y=56$, so $x=16+56+32=104$. Thus, the first player started the game with 104 marbles, the second with 56, and the third with 32.

|  | $1 \underline{0}$ | $2 \underline{O}$ | $3 \underline{O}$ |
| :---: | :---: | :---: | :---: |
| Start |  |  |  |
| After the 1st round |  |  |  |
| After the 2nd round |  |  |  |
| After the 3rd round | 64 | 64 | 64 |

Solution 2: We fill in the table "from bottom to top," that is, from the end of the game to the beginning. We start with 64 in all three final cells.

Since the first two players doubled their number of marbles in the third round,

|  | $\overline{10}$ | $\overline{2}$ | 3 의 |
| :---: | :---: | :---: | :---: |
| Start |  |  |  |
| After the 1st round |  |  |  |
| After the 2nd round | 32 | 32 | $\overline{128}$ |
| After the 3rd round | 64 | 64 | 64 |

each had 32 marbles, and the third player gave 32 to each of them. We deduce that he had $64+32+32=128$ marbles.

|  | $\overline{10}$ | $2 \underline{O}$ | $3^{\underline{O}}$ |
| :---: | :---: | :---: | :---: |
| Start |  |  |  |
| After the 1st round | 16 | $32+16+64=112$ | 64 |
| After the 2nd round | 32 | 32 | 128 |
| After the 3rd round | 64 | 64 | 64 |

The player who lost the second round was the second player. Therefore, the table was

Finally,

|  | $1^{\underline{\underline{o}}}$ | $2^{\underline{\underline{o}}}$ | $3^{\underline{\underline{ }}}$ |
| :---: | :---: | :---: | :---: |
| Start | $16+56+32=104$ | 56 | 32 |
| After the 1st round | 16 | $32+16+64=112$ | 64 |
| After the 2nd round | 32 | 32 | 128 |
| After the 3rd round | 64 | 64 | 64 |

Thus, the first player started the game with 104 marbles, the second with 56, and the third with 32.