$14 \cdot 13$ Find the integer solution to the equation $\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{10!}\right]=1001$.
(24th All-Soviet Union Mathematical Olympiad, 1990)
[Solution] From the given equation, we know that $x>0$. And $x \leqslant 6$!
If $x=6!$, then the original equation becomes
$$
720+360+120+30+6=1001,
$$

which is impossible.
Therefore, $x<6$!.
Let $\quad x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+k \cdot 1!$.

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 the expression for $x$ into the given equation, we get
$$
206 a+41 b+10 c+3 d+k=1001 \text {. }
$$

Since
$$
41 b+10 c+3 d+k \leqslant 201,
$$

it follows that
$$
800 \leqslant 206 a \leqslant 1001,
$$

so
$$
a=4 \text {. }
$$

Thus,
$$
41 b+10 c+3 d+k=177 \text {. }
$$

Since
$$
10 c+3 d+k \leqslant 37,
$$

then
$$
140 \leqslant 41 b \leqslant 177 \text {, }
$$

so
$$
b=4 \text {. }
$$

Thus,
$$
10 c+3 d+k=13 \text {. }
$$

Therefore, we get
$$
c=1, d=1, k=0 \text {. }
$$

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