Let

$$
f(x)=x-\left[\frac{x}{2}\right]-\left[\frac{x}{4}\right]-\left[\frac{x}{6}\right]-\left[\frac{x}{12}\right]
$$

Determine the range of the function.
Since $x=\frac{x}{2}+\frac{x}{4}+\frac{x}{6}+\frac{x}{12}$, therefore

$$
f(x)=\frac{x}{2}-\left[\frac{x}{2}\right]+\frac{x}{4}-\left[\frac{x}{4}\right]+\frac{x}{6}-\left[\frac{x}{6}\right]+\frac{x}{12}-\left[\frac{x}{12}\right]
$$

No number is less than its integer part, so $f(x) \geq 0$. On the other hand, the difference between a number and its integer part is less than one, so $f(x)<4$.

We will show that the range of the function is the interval $[0,4]$.

If $0 \leq t \leq 2$, then

$$
0<\frac{t}{12}<\frac{t}{6}<\frac{t}{4}<\frac{t}{2} \leq 1
$$

thus

$$
\left[\frac{t}{2}\right]=\left[\frac{t}{4}\right]=\left[\frac{t}{6}\right]=\left[\frac{t}{12}\right]=0
$$

so in this case $f(t)=t$.

If $2<t<4$, then

$$
-1<\frac{t-4}{2}<\frac{t-4}{4}<\frac{t-4}{6}<\frac{t-4}{12}<0
$$

thus

$$
\left[\frac{t-4}{2}\right]=\left[\frac{t-4}{4}\right]=\left[\frac{t-4}{6}\right]=\left[\frac{t-4}{12}\right]=-1
$$

and so $f(t-4)=t-4+4=t$.

This shows that for every $t$ such that $0 \leq t<4$, there exists an $x$ such that $f(x)=t$.

Remark. It can generally be proven that if the positive numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ satisfy $\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n}=1$, then the range of the function $f(x)=x-\left[\alpha_{1} x\right]-\left[\alpha_{2} x\right]-\ldots-\left[\alpha_{n} x\right]$ is the interval $[0, n)$.