## Task 1 - 300831

Determine the number of natural numbers that contain each of the digits 1, 2, 3, ..., 9 exactly once and are divisible by 45!
Every natural number that contains each of the digits 1 through 9 exactly once has a cross sum of 45 due to $1+2+3+4+5+6+7+8+9=45$ and is thus divisible by 9.

Of these numbers, the ones whose count is sought are, due to $45=9 \cdot 5$ (and the coprimality of 9 and 5), exactly those divisible by 5. These are precisely the numbers whose last digit is 5, as the digit 0 does not occur according to the problem statement.

To determine the desired count, one must investigate how many different arrangements can be formed from the remaining eight digits:

Using two digits, exactly two arrangements can be formed (e.g., 12, 21).

When adding a third digit, it can stand in the first, second, or third position. Consequently, from each of the two arrangements (e.g., 12, 21), exactly three new ones arise, resulting in a total of $2 \cdot 3=6$ arrangements.

When adding a fourth digit, it can again appear in the first, second, third, or fourth position in each of the already determined six arrangements of three digits. Consequently, with four digits, exactly $2 \cdot 3 \cdot 4=24$ arrangements are possible.

Continuing this reasoning, one concludes that using eight digits, exactly $2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8=40320$ arrangements can be formed.

The desired count is thus 40320.