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
$41 b+10 c+3 d+k \leqslant 201$, it follows that
$$
800 \leqslant 206 a \leqslant 1001,
$$

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 $\quad x=4 \cdot 5!+4 \cdot 4!+3!+2!=584$.