## Task 4 - 320734

Determine the number of all six-digit natural numbers that are divisible by 5 and whose cross sum is divisible by 9!
A natural number $n$ has a digit sum divisible by 9 if and only if $n$ itself is divisible by 9.

Furthermore, $n$ is, due to the coprimality of 5 and 9, divisible by 5 and 9 if and only if $n$ is divisible by 45, i.e., of the form $n=45 \cdot k$ with a natural number $k$.

Since $45 \cdot 2222=99990, 45 \cdot 2223=100035$ and $45 \cdot 22222=999990, 45 \cdot 22223=1000035$, the numbers $45 \cdot k$ with natural $k$ are exactly six-digit for $k=2223, \ldots, 22222$.

Since there are $22222-2222=20000$ values of $k$, the sought number is thus determined to be 20000.