2. When a five-digit number is divided by 100, we get a quotient $k$ and a remainder $o$. For how many five-digit numbers is the sum $k+o$ divisible by 11?
II/2. There are 90000 five-digit numbers (from 10000 to 99999). We can divide them into 900 columns, each containing 100 numbers, based on the value of $k$ from 100 to 999:

$$
\begin{array}{ccccc}
100 \cdot 100+0 & 100 \cdot 101+0 & 100 \cdot 102+0 & \cdots & 100 \cdot 999+0 \\
100 \cdot 100+1 & 100 \cdot 101+1 & 100 \cdot 102+1 & \cdots & 100 \cdot 999+1 \\
\vdots & \vdots & \vdots & \vdots & \vdots \\
100 \cdot 100+99 & 100 \cdot 101+99 & 100 \cdot 102+99 & \cdots & 100 \cdot 999+99
\end{array}
$$

In each column where $k$ is not divisible by 11, we find 9 values of the remainder $o$ such that the sum $k+o$ is divisible by 11. In columns where $k$ is divisible by 11, there are 10 values of the remainder $o$ such that the sum $k+o$ is divisible by 11. In this case, the remainder $o$ can be $0, 11, 22, \ldots, 99$.

We then determine that there are 81 columns where the quotient $k$ is divisible by 11. These are the columns where $k$ has values $110=11 \cdot 10, 121=11 \cdot 11, 132=11 \cdot 12, \ldots, 990=11 \cdot 90$. Therefore, the number of columns where $k$ is not divisible by 11 is $900-81=819$. The sum $k+o$ is divisible by 11 in $81 \cdot 10 + 819 \cdot 9 = 8181$ five-digit numbers.

Finding: For $k$ that is not divisible by 11, there are 9 such values of $o$ such that $k+o$ is divisible by 11 ... 2 points Finding: For $k$ that is divisible by 11, there are 10 such values of $o$ such that $k+o$ is divisible by 11 ... 2 points In 81 cases, $k$ is divisible by 11 ..................................................................................................................... In 819 cases, $k$ is not divisible by 11 .................................................................................................... The sum $k+o$ is divisible by 11 in 8181 five-digit numbers .........................................................