A number is divisible by 45 if and only if it is divisible by both 5 and 9. Therefore, the digit in the units place must be 0 or 5, and the sum of its digits must be a multiple of nine.

First, we determine the number of the desired numbers that have the digit 0 in the units place. We denote these numbers as $\overline{1 A 2 B 30}$, and their digit sum is then $6 + A + B$. For this sum to be a multiple of nine, and considering that the unknowns $A$ and $B$ represent digits from 0 to 9, the digit sum can be either 9 or 18. In the first case, $A + B = 3$, and in the second case, $A + B = 12$. The following tables show how many pairs of digits give a sum of 3 and 12, respectively:

| $A$ | 3 | 2 | 1 | 0 |
| :---: | :---: | :---: | :---: | :---: |
| $B$ | 0 | 1 | 2 | 3 |


| $A$ | 9 | 8 | 7 | 6 | 5 | 4 | 3 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $B$ | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

Thus, there are $4 + 7 = 11$ numbers of the form $\overline{1 A 2 B 30}$ that are divisible by 45.

Next, we determine the number of the desired numbers that have the digit 5 in the units place. We denote these numbers as $\overline{1 C 2 D 35}$, and their digit sum is then $11 + C + D$. Similarly to the previous part of the problem, we find that either $C + D = 7$ or $C + D = 16$. We again create tables:

| $C$ | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $D$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |


| $C$ | 9 | 8 | 7 |
| :---: | :---: | :---: | :---: |
| $D$ | 7 | 8 | 9 |

Thus, there are $8 + 3 = 11$ numbers of the form $\overline{1 C 2 D 35}$ that are divisible by 45. In total, there are $11 + 11 = 22$ numbers that meet the criteria.

Note. Students can also initially divide the desired numbers into groups with digit sums of 9, 18, and 27. In the group with a digit sum of 9, the units digit can only be 0, in the group with a digit sum of 27, the units digit can only be 5, and in the group with a digit sum of 18, both of these digits can be in the units place.

Evaluation. 1 point for the condition of divisibility by 45; 1 point for dividing the desired numbers into groups; 4 points for correctly determining the numbers in each group.
How many six-digit natural numbers exist that have the digit 1 in the thousands place, the digit 2 in the hundreds place, and the digit 3 in the tens place, and are divisible by 45 without a remainder?

(L. Simuinek)