How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8,9$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating?
For the number to be divisible by $5$ it must end with $0$ (or $5$, but we don't have it in the given problem). So we have $V_6^4$ numbers which are divisible by $5$ but how can we exclude those which aren't divisible by $3$?

 $120$


We must select the $5$ digits which must appear, although we know the $0$ must appear. So in reality we must only select $4$ digits out of $6$ possibilities.
Notice the sum of all those numbers is a multiple of $3$ ($1+2+6+7+8+9 = 33)$. Therefore the condition is that the two numbers we do not pick add up to a multiple of $3$.
There are $4$ possible ways to do this if these numbers are $1$ and $2\bmod 3$ and there is one option when both unselected numbers are multiples of $3$.
We conclude there are $5$ valid selections for the $5$ digits that appear. For each configuration there are $4!$ rearrangements.
We conclude the answer is $5\times 4! = 120$
