4. 174 Find the integer solutions of the following equation
$$\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{10!}\right]=1001$$
[Solution] From the equation, we know that $x$ is a natural number not exceeding 1001, so $x<6$!, and thus for the terms $\left[\frac{x}{n!}\right]$ when $n \geqslant 6$, they can be omitted. Each $x<6!$ can be uniquely represented as
$$x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+k \cdot 1!$$
where $a, b, c, d, k$ are non-negative integers, and
$$a \leqslant 5, b \leqslant 4, c \leqslant 3, d \leqslant 2, k \leqslant 1$$

Substituting this expression for $x$ into the equation, we get
$$206 a+41 b+10 c+3 d+k=1001$$

Since
$$\begin{array}{l}
41 b+10 c+3 d+k \leqslant 201, \text{ it follows that } \\
800 \leqslant 206 a \leqslant 1001,
\end{array}$$

which means $a=4$, implying
$$41 b+10 c+3 d+k=177$$

Continuing with similar discussions, we get
$$b=4, c=d=1, k=0$$

Therefore, we have $\quad x=4 \cdot 5!+4 \cdot 4!+3!+2!=584$.