## Task A-1.5.

On a board, there are 2023 different real numbers. If each number on the board (simultaneously) is replaced by the sum of all the other numbers, the 2023 numbers on the board will be the same as at the beginning.

What values can the product of all the numbers on the board take at some point?
## First Solution.

Let the numbers initially written on the board be denoted by $x_{1}, x_{2}, \ldots, x_{2023}$, and let $s=x_{1}+x_{2}+\ldots+x_{2023}$.

Since after the first swap the numbers on the board are the same as the initial numbers, it follows that after any other swap, the numbers on the board will also be the initial numbers.

Therefore, the product of all numbers on the board will always be the same as the product of the initial numbers on the board. The same statement holds for the sum of the numbers on the board.

After the first swap, the numbers on the board will be $s-x_{1}, s-x_{2}, \ldots, s-x_{2023}$.

Since these are the same numbers that were initially written on the board, their sum is also equal to $s$, i.e.,

$$
s=\left(s-x_{1}\right)+\left(s-x_{2}\right)+\ldots+\left(s-x_{2023}\right)=2023 s-\left(x_{1}+x_{2}+\ldots+x_{2023}\right)=2022 s . \quad 1 \text { point }
$$

From the above equation, it follows that $s=0$.

Therefore, the numbers on the board after the first swap are $-x_{1},-x_{2}, \ldots,-x_{2023}$.

Since these are the same numbers that were initially on the board, the product of all numbers before and after the swap will be the same, i.e.,

$$
x_{1} \cdot x_{2} \cdot \ldots \cdot x_{2023}=\left(-x_{1}\right) \cdot\left(-x_{2}\right) \cdot \ldots \cdot\left(-x_{2023}\right)=-x_{1} \cdot x_{2} \cdot \ldots \cdot x_{2023}
$$

From the above equality, we have $x_{1} \cdot x_{2} \cdot \ldots \cdot x_{2023}=0$.

Thus, the product of all numbers on the board at any time will always be 0.