$$
6 \cdot 13 \text { For } f(x)=[x]+[2 x]+\left[\frac{5}{3} x\right]+[3 x]+[4 x] \text {, }
$$

When the real number $x$ varies in $0 \leqslant x \leqslant 100$, try to find the number of distinct integer values that can be obtained.

[Solution] First, for $0 \leqslant x \leqslant 100, x$ can be uniquely written in the form $x=3 n+z$, where $n$ is a non-negative real number, and $0 \leqslant z<3$. Therefore, $f(x)=f(3 n+z)=35 n+f(z)$.

Second, divide $[0,99)$ into 33 small intervals $[n, n+3), n=0,3,6,9, \cdots, 96$. When $x$ is restricted to each small interval, the number of distinct real values of $f(x)$ is the same. Therefore, we only need to consider the case when $x \in[0,3)$. $f$ is a non-decreasing step function, so a sufficient condition for $x$ to be a jump point of $f$ is that $x$ is at least one of $[x],[2 x],\left[\frac{5}{3} x\right],[3 x]$, $[4 x]$. Therefore, in $[0,3)$, $f$ has 22 such jump points, that is,
$$
\begin{array}{l}
0, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, 1, \frac{6}{5}, \frac{5}{4}, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}, \frac{7}{4}, \frac{9}{5}, 2, \frac{9}{4}, \frac{7}{5}, \\
\frac{12}{5}, \frac{5}{2}, \frac{8}{3}, \frac{11}{4} .
\end{array}
$$

Therefore, when $x \in[0,3)$, $f$ has 22 distinct integer values. Thus, in $[0,99)$, $f$ has $22 \times 33=726$ distinct integer values.

Finally, consider the case when $x \in[99,100]$. The number of distinct integer values of $f$ in $[99,100]$ is the same as the number of distinct integer values of $f$ in $[0,1]$. The jump points of $f$ in $[0,1]$ are $0, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, 1$, a total of 8. Therefore, $f$ has 8 distinct integers in $[0,1]$.
Thus, $f$ has a total of $726+8=734$ different integers in $[0,100]$.