$14 \cdot 38$ When $0 \leqslant x \leqslant 100$, find the number of different integers taken by the function
$$
f(x)=[x]+[2 x]+\left[\frac{5 x}{3}\right]+[3 x]+[4 x]
$$

(29th International Mathematical Olympiad Candidate Problem, 1988 Asia Pacific Mathematical Olympiad, 1993)
[Solution 1] (1) If $x=3 k+\alpha, k$ is a non-negative integer, $0 \leqslant \alpha<1$, then
$$
\begin{array}{l}
f(x)=3 k+[6 k+2 \alpha]+\left[5 k+\frac{5}{3} \alpha\right]+[9 k+3 \alpha]+[12 k+4 \alpha] \\
=35 k+\left[\frac{5}{3} \alpha\right]+[2 \alpha]+[3 \alpha]+[4 \alpha] \\
0 \text {, if } 0 \leqslant \alpha<\frac{1}{4} \text {; } \\
1 \text {, if } \frac{1}{4} \leqslant \alpha<\frac{1}{3} \text {; } \\
\text { 2, if } \frac{1}{3} \leqslant \alpha<\frac{1}{2} \text {; } \\
=35 k+\left\{4 \text {, if } \frac{1}{2} \leqslant \alpha<\frac{3}{5} ;\right. \\
5 \text {, if } \frac{3}{5} \leqslant \alpha<\frac{2}{3} \text {; } \\
\text { 6, if } \frac{2}{3} \leqslant \alpha<\frac{3}{4} \text {; } \\
\text { 7, if } \frac{3}{4} \leqslant \alpha<1 \text {. } \\
\end{array}
$$

For a fixed $k, f(x)$ has 7 different values, i.e., when $\alpha$ changes from one interval to another, the value of $f(x)$ changes once
$$
0 \quad \frac{1}{4} \quad \frac{1}{3} \quad \frac{1}{2} \frac{3}{5} \quad \frac{2}{3} \quad \frac{3}{4} \quad 1
$$
(2) If $x=3 k+1+\alpha, \quad k$ is a non-negative integer, $0 \leqslant \alpha<1$, then

For a fixed $k, f(x)$ has 8 different values, when $\alpha$ changes from one interval to another, the value of $f(x)$ changes once.
(3) If $x=3 k+2+\alpha, k$ is a non-negative integer, $0 \leqslant \alpha<1$. Then
$$
f(x)=35 k+23+[2 \alpha]+[3 \alpha]+[4 \alpha]+\left[\frac{5}{3} \alpha+\frac{1}{3}\right]
$$
$\left(0\right.$, if $0 \leqslant \alpha<\frac{1}{4}$;
1, if $\frac{1}{4} \leqslant \alpha<\frac{1}{3}$;
2, if $\frac{1}{3} \leqslant \alpha<\frac{2}{5}$;
3, if $\frac{2}{5} \leqslant \alpha<\frac{1}{2}$;
5, if $\frac{1}{2} \leqslant \alpha<\frac{2}{3} ;$
6, if $\frac{2}{3} \leqslant \alpha<\frac{3}{4}$;
7, if $\frac{3}{4} \leqslant \alpha<1$.

For a fixed $k, f(x)$ has 7 different values, when $\alpha$ changes from one interval to another, the value of $f(x)$ changes once

When $x=100$, i.e., for the second case, $k=33, \alpha=0$, $f(x)$ has only one value $35 \cdot 33+11$

While when $0 \leqslant x \leqslant 99$, in the first case, $k$ can take 34 values from 0 to 33, in the second case, $k$ can take 33 values from 0 to 32, and in the third case, $k$ can take 33 values from 0 to 32.
So when $0 \leqslant x \leqslant 100$, $f(x)$ can take
$$
34 \cdot 7+33 \cdot 8+33 \cdot 7+1=734
$$

different integer values.
[Solution 2] Let $x=3 n+r, n \geqslant 0,0 \leqslant r<3$, then
$$
\begin{aligned}
f(x) & =f(3 n+r) \\
& =[3 n+r]+[6 n+2 r]+\left[5 n+\frac{5 r}{3}\right]+[9 n+3 r]+[12 n+4 r] \\
& =3 n+6 n+5 n+9 n+12 n+[r]+[2 r]+\left[\frac{5 r}{3}\right]+[3 r]+[4 r] \\
& =35 n+f[r] . \\
& \text { This way, for each } n \geqslant 0 \text {, in the interval }[n, n+3) \text { the function } f(x) \text { gives the same }
\end{aligned}
$$

This way, for each $n \geqslant 0$, in the interval $[n, n+3)$ the function $f(x)$ gives the same number of different integer values.
For this, it is only necessary to consider the interval $[0,3)$.
Note that $f(x)$ is a step function, and its value jumps at least at one of the points $[x],[2 x]$ , $\left[\frac{5 x}{3}\right],[3 x],[4 x]$. Clearly, $f(x)$ jumps 22 times in the interval $[0,3)$, i.e., at
$0, \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}$,
$1, \frac{6}{5}, \frac{5}{4}, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}, \frac{7}{4}, \frac{9}{5}$,
$2, \frac{9}{4}, \frac{7}{3}, \frac{12}{5}, \frac{5}{2}, \frac{8}{3}, \frac{11}{4}$

Thus, the number of different integer values that the function $f(x)$ takes in the interval $[0,3)$ is 22.
Therefore, the number of different integer values in the interval $[0,99)$ is
$$
22 \cdot 33=726 \text {. }
$$

Similarly, the number of different integer values that the function takes in the interval $[99,100]$ is the same as in $[0,1]$, and in $[0,1]$ the function has 8 jumps, i.e., at
$0, \quad \frac{1}{4}, \frac{1}{3}, \frac{1}{2}, \frac{3}{5}, \frac{2}{3}, \frac{3}{4}, 1$

Thus, the number of different integer values that the function $f(x)$ takes in the interval $[0,1]$ is 8, and therefore in $[99,100]$ the number of different integer values that $f(x)$ takes is also 8.
So, the number of different integer values that $f(x)$ takes in the interval $[0,100]$ is