## Task 1 - 030831

What is the smallest eight-digit number that consists of all different digits and is divisible by 36? Justify that it is the smallest such number!
A number is divisible by 36 if it is divisible by 9 and 4.

A number is divisible by 9 if the sum of its digits is divisible by 9.

A number is divisible by 4 if the number formed by its last two digits is divisible by 4.

The number should consist of 8 different digits, and there are 10 digits (0-9). Therefore, 2 digits must be omitted so that the sum of the remaining digits is divisible by 9. For this, the pairs of numbers $(0 ; 9),(1 ; 8),(2 ; 7),(3 ; 6)$, and $(4 ; 5)$ are possible. This gives the following minimal possible numbers:

$$
\begin{gathered}
(0 ; 9)-12345678 \quad(1 ; 8)-20345679 \quad(2 ; 7)-10345689 \\
(3 ; 6)-10245789 \quad(4 ; 5)-10236789
\end{gathered}
$$

The pair (4;5) allows for the smallest number, but the digits must be arranged so that the number is divisible by 4. To make the number as small as possible, the larger digits should be placed further back. The number formed by the largest possible 2 digits that is divisible by 4 is then 96. Therefore, only the last 4 digits need to be rearranged.

This results in the number 10237896.