1B. Alexandar, Bojan, Vasil, and Goran played games with chips. They played a total of four games, and each of them won exactly one game, in the order given above. At the end of each game and before the start of the next game, the winner of the last game gave each of the other players as many chips as they had at that moment. After all the games were played, each player had exactly 48 chips. How many chips did each player have before the start of the first game?
Solution. First method. After all the games were played, each of the players had exactly 48 chips. This means that $(48,48,48,48)$ are the number of chips after all the games played by Alexander, Bojan, Vasil, and Goran, respectively. The winner of the fourth game was Goran, which means he doubled the chips of the other players, so before the start of the fourth game, the players had

$$
48: 2=24,48: 2=24,48: 2=24 \text { and } 48+24+24+24=120
$$

chips respectively, i.e., $(24,24,24,120)$. The winner of the third game was Vasil, which means he doubled the chips of the other players, so before the start of the third game, the players had

$$
24: 2=12,24: 2=12,24+12+12+60=108 \text { and } 120: 2=60
$$

chips respectively, i.e., $(12,12,108,60)$. The winner of the second game was Bojan, which means he doubled the chips of the other players, so before the start of the second game, the players had

$$
12: 2=6,12+6+54+30=102,108: 2=54 \text { and } 60: 2=30
$$

chips respectively, i.e., $(6,102,54,30)$. The winner of the first game was Alexander, which means he doubled the chips of the other players, so before the start of the first game, the players had

$$
6+51+27+15=99,102: 2=51,54: 2=27 \text { and } 30: 2=15
$$

chips respectively, i.e., $(99,51,27,15)$.

Thus, before the start of the first game, Alexander had 99, Bojan had 51, Vasil had 27, and Goran had 15 chips.

Second method. Let $a, b, v, g$ be the number of chips before the game for Alexander, Bojan, Vasil, and Goran, respectively.

In the first game, the winner is Alexander, so before the start of the second game, Alexander had $a-b-v-g$, Bojan had $2 b$, Vasil had $2 v$, and Goran had $2 g$ chips.

In the second game, the winner is Bojan, so before the start of the third game, Alexander had $2(a-b-v-g)$, Bojan had

$$
2 b-(a-b-v-g)-2 v-2 g=3 b-a-v-g
$$

Vasil had $4 v$ and Goran had $4 g$ chips.

In the third game, the winner is Vasil, so before the start of the fourth game, Alexander had $4(a-b-v-g)$, Bojan had $2(3 b-a-v-g)$, Vasil had

$$
4 v-2(a-b-v-g)-(3 b-a-v-g)-4 g=7 v-a-b-g
$$

and Goran had $8 g$ chips.

In the fourth game, the winner is Goran, so after all the games were played, Alexander had $8(a-b-v-g)$, Bojan had $4(3 b-a-v-g)$, Vasil had $2(7 v-a-b-g)$, and Goran had

$$
8 g-4(a-b-v-g)-2(3 b-a-v-g)-(7 v-a-b-g)=15 g-a-b-v
$$

chips.

Since after all the games were played, each had exactly 48 chips, we have

$$
\left\{\begin{array} { l } 
{ 8 ( a - b - v - g ) = 4 8 } \\
{ 4 ( 3 b - a - v - g ) = 4 8 , } \\
{ 2 ( 7 v - a - b - g ) = 4 8 , } \\
{ 1 5 g - a - b - v = 4 8 }
\end{array} \quad \text { i.e., } \quad \left\{\begin{array}{l}
a-b-v-g=6 \\
3 b-a-v-g=12 \\
7 v-a-b-g=24 \\
15 g-a-b-v=48
\end{array}\right.\right.
$$

Also, the total number of chips is $4 \cdot 48=192$, so we have

$$
a+b+v+g=192
$$

Solving the system of the last five equations, we get $a=99$, $b=51$, $v=27$, and $g=15$. Thus, Alexander had 99 chips, Bojan had 51 chips, Vasil had 27 chips, and Goran had 15 chips before the start of the first game.

2AB. Given the system of linear equations

$$
\left\{\begin{array}{l}
(a+3) x-2(a+3) y=12-3 a \\
(a+3) x+(a+3) y=-1-a
\end{array}\right.
$$

where $a \neq-3$ is a parameter that does not depend on $x$ and $y$. Determine all integer values of the parameter $a$ for which the solutions of the system are negative numbers.