Solve the following equation:

$$
x-1=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{6}\right]
$$
The right side of the equation is an integer, so $x$ can only be an integer. If $r$ represents the remainder when $x$ is divided by 6, then $x-r$ is divisible by 6, that is, $x=6k+r$, where $k$ is an integer. In this case, (1) transforms into:

$$
6k+r-1=\left[3k+\frac{r}{2}\right]+\left[2k+\frac{r}{3}\right]+\left[k+\frac{r}{6}\right]
$$

It is known that if $n$ is an integer, then $[n+t]=n+[t]$. Using this on the right side of (2), after rearrangement,

$$
r-1=\left[\frac{r}{2}\right]+\left[\frac{r}{3}\right]+\left[\frac{r}{6}\right] \text{ is obtained. }
$$

Thus, the integer solutions of equation (1) are exactly those integers for which the remainder when divided by 6 satisfies (3). Since the possible values of $r$ are $0,1,2,3,4$ and $5$, checking (3) involves six substitutions. It can finally be seen that (3) is satisfied exactly when the value of $r$ is $1,2,3$ or 4.

Therefore, the solutions of equation (1) are the numbers of the form $6k+1, 6k+2, 6k+3$, and $6k+4$, where $k$ is any integer.