8. The smallest solution to the equation $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x$ is $\qquad$ (where, $[x]$ denotes the greatest integer not exceeding $x$).
Let $x=42 a+b, 0 \leqslant b \leqslant 41$, substituting into the equation we get $\left[\frac{b}{2}\right]+\left[\frac{b}{3}\right]+\left[\frac{b}{7}\right]=a+b$, then we have $a \geqslant \frac{b-1}{2}+\frac{b-2}{3}+\frac{b-6}{7}-b=\frac{-b-85}{42} \geqslant-3$, equality holds when $a=-3, b=41$. Therefore, the minimum solution for $x$ is $42 \times(-3)+41=-85$.