Example 3 Find the number of different integer values taken by the function $f(x)=[x]+[2 x]+[3 x]+\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]$ on $0 \leqslant x \leqslant 100$.

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$f(x)$ is a step function, and its value needs to increase, it must jump up at one of the points $[x],[2 x],[3 x],\left[\frac{x}{2}\right]$, $\left[\frac{x}{3}\right]$. Therefore, when it increases, the value of $x$ must be a multiple of 1, or $\frac{1}{2}$, or $\frac{1}{3}$, or 2, or 3, and within the interval $[0,6)$, there are a total of $0, \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 1, \frac{4}{3}, \frac{3}{2}, \frac{5}{3}, 2, \frac{7}{3}, \frac{5}{2}, \frac{8}{3}, 3, \frac{10}{3}, \frac{7}{2}, \frac{11}{3}, 4, \frac{13}{3}, \frac{9}{2}, \frac{14}{3}, 5, \frac{16}{3}$, $\frac{11}{2}, \frac{17}{3}$, these 24 numbers are multiples of 1, or $\frac{1}{2}$, or $\frac{1}{3}$, or 2, or 3. Thus, the number of different values $f(x)$ takes in the interval $[0,6)$ is 24, and the number of different values it takes in $[0,96)$ is $24 \times 16=384$. Furthermore, the number of values $f(x)$ takes in $[96,100]$ is the same as the number of values it takes in $[0,4]$, and in $[0,4]$, $f(x)$ has 17 jumps, so the number of different values $f(x)$ takes in $[0,100]$ is $384+17=401$.