1. Let $n!=1 \times 2 \times \cdots \times n,[x]$ denote the greatest integer not exceeding the real number $x$. Then the positive integer solution to the equation
$$
\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{10!}\right]=3468
$$

is $\qquad$
When $x=2020$,
the left side of the equation
$$
\begin{array}{l}
=2020+1010+336+84+16+2 \\
=3468=\text { right side. }
\end{array}
$$

Since the left side function is an increasing function, $x=2020$ is the unique solution of the equation.