8. For any real number $x$, $[x]$ represents the greatest integer not exceeding $x$. When $0 \leqslant x \leqslant 100$, the range of $f(x)=[3 x]+[4 x]+[5 x]+[6 x]$ contains $\qquad$ different integers.
8. 1201 .

Solution: Since $f(x+1)=f(x)+18$, divide $[0,100]$ into intervals $[0,1),[1,2), \cdots,[99,100)$ and $\{100\}$.

Notice that $f(x)$ has a "jump" at fractions with denominators of $3,4,5$ or 6 ( $\frac{1}{2}$ can also be considered as having a denominator of 4, because $\frac{1}{2}=$ $\left.\frac{2}{4}\right)$, and there are 11 different fractions with denominators of $3,4,5$ or 6 in $[0,1)$, so $[0,1)$ is divided into 12 left-closed right-open subintervals, each corresponding to a different value of $f(x)$, meaning $f(x)$ takes 12 different values on $[0,1)$. Therefore, when $0 \leqslant x \leqslant 100$, the range of $f(x)$ contains 1201 different integers.