$1 \cdot 11$ For $0 \leqslant x \leqslant 100$, find the number of different integer values taken by the function $f(x)=[x]+[2 x]+\left[\frac{5 x}{3}\right]+ [3 x]+[4 x]$.

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[Solution] Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ represent the sets of all discontinuity points of the functions $[x],[2 x],[3 x]$, $[4 x]$, and $\left[\frac{5 x}{3}\right]$, respectively. It is easy to see that $A_{1} \subset A_{2} \subset A_{4}$ and
$$
\begin{array}{l}
A_{3}=\left\{\left.\frac{n}{3} \right\rvert\, n=1,2, \cdots, 300\right\}, \\
A_{4}=\left\{\left.\frac{n}{4} \right\rvert\, n=1,2, \cdots, 400\right\}, \\
A_{5}=\left\{\left.\frac{3 n}{5} \right\rvert\, n=1,2, \cdots, 166\right\} .
\end{array}
$$

From this, we can obtain
$$
\begin{aligned}
A_{3} \cap A_{4} & =\{n \mid n=1,2, \cdots, 100\}, \\
A_{3} \cap A_{5} & =A_{4} \cap A_{5}=A_{3} \cap A_{4} \cap A_{5} \\
& =\{3 n \mid n=1,2, \cdots, 33\} .
\end{aligned}
$$

By the principle of inclusion-exclusion, the number of discontinuity points of $f(x)$ is
$$
\begin{aligned}
& \left|A_{3}\right|+\left|A_{4}\right|+\left|A_{5}\right|-\left|A_{3} \cap A_{4}\right|-\left|A_{3} \cap A_{5}\right|-\left|A_{4} \cap A_{5}\right| \\
& +\left|A_{3} \cap A_{4} \cap A_{5}\right| \\
= & 300+400+166-100-33-33+33=733 .
\end{aligned}
$$

Therefore, the number of different integer values that $f(x)$ can take is 734.