## Task 2 - 330712

A six-digit natural number should, when read from left to right, have the digits $3, a, 3, b, 2, c$.

Determine all possibilities for choosing the digits $a, b, c$ such that the specified number is divisible by 90!
Since 9 and 10 are coprime, the mentioned number is exactly divisible by 90 if and only if it is divisible by 9 and by 10. It is exactly divisible by 10 if $c=0$ (1). It is, if (1) holds, also exactly divisible by 9 if the cross sum $3+a+3+b+2+c=8+a+b$ is divisible by 9, i.e., exactly when $a+b$ leaves a remainder of 1 upon division by 9.

Given $0 \leq a \leq 9$ and $0 \leq b \leq 9$, this is exactly the case for $a+b=1$ and $a+b=10$. For these, there are exactly the following possibilities (2):

$$
\begin{array}{l|lllllllllll}
\mathrm{a} & 0 & 1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\mathrm{~b} & 1 & 0 & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1
\end{array}
$$

In (1) and (2), all possibilities for the mentioned choice of $a, b, c$ are thus given.