You can reduce the possibilities quite quickly. Clearly the final digit has to be $0$, and then divisibility by $10$ is assured as is divisibility by $9$.

You can only have divisibility by $5$ if $5$ is in the fifth place, and divisibility by $2$ only works if the digits in even places are even.

Divisibility by $8$ means the digits in the $7, 8$ places are restricted to $32, 72, 16, 96$ because $0, 5$ are taken and there are no two consecutive even digits.

Divisibility by $4$ has the third and fourth places taken by a choice of $12, 32, 72, 92, 16, 36, 76, 96$

This means that $4, 8$ occupy places $2$ and $6$ in some order.

Then the test for divisibility by $3$ can reduce possibilities still further, and is simply applied, leaving divisibility by $7$ as the final arbiter of the remaining options.