6. Find at least one solution to the equation

$$
567 x^{3}+171 x^{2}+15 x-777 \ldots 7555 \ldots 5333 \ldots 3=0
$$

where the constant term contains 2023 sevens, 2023 fives, and 2023 threes.
Answer. $\quad 111 \ldots 1_{w} 2023$.

Solution. Let the number whose decimal representation consists of 2023 ones be denoted by $x$ and note the following property of this number:

$$
111 \ldots 1_{\longleftarrow} 2023=x, \quad 9 x+1=10^{2023}
$$

Represent the number $777 \ldots 7555 \ldots 5333 \ldots 3$ as follows:

$$
\begin{gathered}
777 \ldots 7555 \ldots 5333 \ldots 3=3 x+5 x * 10^{2023}+7 x * 10^{2 * 2023}= \\
=3 x+5 x *(9 x+1)+7 x *(9 x+1)^{2}=567 x^{3}+171 x^{2}+15 x
\end{gathered}
$$

Therefore, the number $x$ is a root of the original equation.

Criteria. A correct answer obtained with justification - 7 points. A solution based on the study of a similar equation with a smaller coefficient, for example, such as

$567 x^{3}+171 x^{2}+15 x-775533=0$ and then generalizing this case. If from the observation that $x=11$ is a solution of this equation without any additional reasoning, a conclusion is made about the solution of the original equation - 3 points. The expression $9 x+1=10^{2023}$ or its equivalent is obtained, but the solution is not completed, the answer is not obtained - 2 points.

Attempts to study the structure of the free coefficient, some representation of this number using the number 111...1 (2023 ones) is provided, the solution is not completed, the answer is not obtained - 1 point.