How many positive integer solutions does the following equation have?

$$
\left[\frac{x}{10}\right]=\left[\frac{x}{11}\right]+1
$$
Every sought $x$ value can be uniquely written in the form $x=11 \cdot k+m$, where $k$ is a non-negative integer, and $m$ takes one of the values 0, $1, \ldots, 10$. Then the equation is:

$$
\left[\frac{11 \cdot k+m}{10}\right]=\left[\frac{11 \cdot k+m}{11}\right]+1
$$

Based on the interpretation of the integer part function,

$$
\left[\frac{11 \cdot k+m}{10}\right]=\left[k+\frac{k+m}{10}\right]=k+\left[\frac{k+m}{10}\right]
$$

and

$$
\left[\frac{11 \cdot k+m}{11}\right]=k
$$

Thus,

$$
k+\left[\frac{k+m}{10}\right]=k+1
$$

which implies

$$
\left[\frac{k+m}{10}\right]=1
$$

The solutions to this are given by $10 \leq k+m \leq 19$. That is, $10-m \leq k \leq 19-m$, where $m=0,1, \ldots, 10$ can be.

Thus, for each value of $m$, there are ten possible values of $k$, and different $(k, m)$ pairs determine different $x$ values. This means that the equation has $10 \cdot 11=110$ solutions.