## Task A-1.1.

Around a round table, there are ten chairs numbered from 1 to 10 (with chairs 1 and 10 being adjacent), and each chair is occupied by a knight. Each knight initially has an even number of gold coins. Simultaneously, each knight gives half of their gold coins to their left neighbor and half to their right neighbor. After this, the knight on chair 1 has 22 gold coins, and each subsequent knight has two more, up to the knight on chair 10 who has 40 gold coins.

How many gold coins did the knight who ends up with 36 gold coins have at the beginning?
## First Solution.

Let's introduce the notation such that $2 x_{1}, 2 x_{2}, \ldots, 2 x_{10}$ represent the number of gold coins that the knights on seats $1, 2, \ldots, 10$ had at the beginning. We need to find $2 x_{8}$.

We have the system:

$x_{10} + x_{2} = 22, x_{1} + x_{3} = 24, x_{2} + x_{4} = 26, x_{3} + x_{5} = 28, \ldots, x_{8} + x_{10} = 38, x_{9} + x_{1} = 40$.

By substitution, we get:

$$
\begin{aligned}
x_{10} & = 38 - x_{8} \\
x_{2} & = 22 - x_{10} = 22 - (38 - x_{8}) = x_{8} - 16 \\
x_{4} & = 26 - x_{2} = 26 - (x_{8} - 16) = 42 - x_{8} \\
x_{6} & = 30 - x_{4} = 30 - (42 - x_{8}) = x_{8} - 12 \\
x_{8} & = 34 - x_{6} = 34 - (x_{8} - 12) = 46 - x_{8}
\end{aligned}
$$

From the last equation, we get $2 x_{8} = 46$.

The knight who ends up with 36 gold coins had 46 gold coins at the beginning.