8. -85 .

Let $x=42 p+q(p, q \in \mathbf{Z}, 0 \leqslant q \leqslant 41)$.
Substituting $x$ into the original equation, we get
$$
p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q .
$$

For each different $q$, a unique ordered pair $(p, q)$ is determined, thus, $x$ is also unique.
$$
\begin{array}{l}
\text { Also, } p=\left[\frac{q}{2}\right]+\left[\frac{q}{3}\right]+\left[\frac{q}{7}\right]-q \\
\geqslant \frac{q-1}{2}+\frac{q-2}{3}+\frac{q-6}{7}-q=-\frac{q+85}{42} \\
\geqslant-3,
\end{array}
$$

When and only when $q=41$, $p=-3$, at this time, $x$ is the smallest solution -85 .
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. The smallest solution to the equation $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{7}\right]=x$ is $\qquad$