17.1.7 * Find the integer solution to the equation $\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{10!}\right]=1001$.
Obviously, $x$ is a positive integer. Since $x+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{10!}\right]=1001$, it follows that $x \leqslant 1001 < 6!$. The equation becomes
$$
x+\left[\frac{x}{2!}\right]+\left[\frac{x}{3!}\right]+\left[\frac{x}{4!}\right]+\left[\frac{x}{5!}\right]=1001.
$$

Let $x=a \cdot 5!+y, 0 \leqslant a \leqslant 5, 0 \leqslant y < 5!$. Similarly, $y=b \cdot 4!+z, 0 \leqslant z < 4!$. Continuing in this manner, we get $x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+e$, where $0 \leqslant a \leqslant 5$, $0 \leqslant b \leqslant 4$, $0 \leqslant c \leqslant 3$, $0 \leqslant d \leqslant 2$, $0 \leqslant e \leqslant 1$. Substituting into (1) yields
$$
206 a+41 b+10 c+3 d+e=1001.
$$

Since $41 b+10 c+3 d+e \leqslant 201$, it follows that $800 \leqslant 206 a \leqslant 1001$, i.e., $a=4$. Thus, we have $41 b+10 c+3 d+e=177$. Continuing with similar discussions, we get $b=4$, $c=d=1$, $e=0$. Therefore, the integer solution to the equation is $x=4 \cdot 5!+4 \cdot 4!+3!+2!=584$.