3. How many five-digit natural numbers have the sum of the quotient and remainder when divided by 100 divisible by 11?
First solution.

When dividing a five-digit number $n$ by 100, the first three digits of $n$ form the quotient $a$, and the last two digits form the remainder $b$. Then $100 \leq a \leq 999$ and $0 \leq b \leq 99$.

The number of $a$ is $999-99=900$, that is, the number $a$ can be chosen in $9 \cdot 10 \cdot 10=900$ ways.

For each possibility for the number $a$, we observe and count the possibilities for the number $b$.

| $a$ | $b$ |  |
| :---: | :---: | :--- |
| 100 | $10,21, \ldots, 98$ | 9 ways |
| $\ldots$ |  | $\ldots$ |
| 109 | $1,12, \ldots, 89$ | 9 ways |
| 110 | $0,11,22, \ldots, 99$ | 10 ways |
| 111 | $10,21, \ldots, 98$ | 9 ways |
| $\ldots$ |  | $\ldots$ |
| 120 | $1,12, \ldots, 89$ | 9 ways |
| 121 | $0,11,22, \ldots, 99$ | 10 ways |
|  |  |  |

For the sum $a+b$ to be divisible by 11, for each of the 900 choices of $a$, the number $b$ can be chosen in at least as many ways as the integer part of the quotient of 100 and 11, that is, in 9 ways.

When $a \in\{110,121,132, \ldots, 990\}$, that is, when $a$ is divisible by 11, the number $b$ can be chosen in one more, the tenth way.

The number of $a$ that belong to the set $\{110,121,132, \ldots, 990\}$ is as many as the integer part of the quotient of 900 and 11, that is, 81.

The number of desired numbers $n$ is

$$
900 \cdot 9+81=8100+81=8181
$$

##