5 . Find the integer solution to the equation
$$
\left[\frac{x}{1!}\right]+\left[\begin{array}{c}
x \\
2!
\end{array}\right]+\cdots+\left[\frac{x}{10!}\right]=1001
$$
5. From the old equation, we know that $x$ is a positive integer not exceeding 1001, so $x<6$!. Therefore, the last five terms on the left side of the equation can be removed. Each positive integer $x<6$! can be uniquely expressed as
$$
x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+e,
$$

where $a$, $b$, $c$, $d$, $e$ are non-negative integers, and $a \leqslant 5$, $b \leqslant 4$, $c \leqslant 3$, $d \leqslant 2$, $e \leqslant 1$. Substituting (1) into the known equation, we get
$$
\begin{array}{l}
206 a+41 b+10 c+3 d+e=1001 . \\
\text { Since } 41 b+10 c+3 d+e \leqslant 201,
\end{array}
$$

it follows that $800 \leqslant 206 a \leqslant 1001$. This implies $a=4$. This also means
$$
41 b+10 c+3 d+e=177 .
$$

By similar reasoning, we get $b=4$, $c=d=1$, $e=0$. Therefore, $x=4 \times 5!+4 \times 4!+3!+2!=584$.