<s>[INST] 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 $3$?

 $120$

 [/INST] $0$ is last digit. Note that there are only 2 numbers from the set whose mod 3 is 1.
Case 1: Pick $0$ $x$ s.t. $x \mod 3 = 1$: Not possible
Case 2: Pick exactly 1 digit x s.t. $x \mod 3 = 1$. We must pick exactly 1 digit x s.t. $x \mod 3 = 2$. Remaining are filled with $x \mod 3 = 0$.
Sum = $2 * 2 * (4!) = 96$
case 3: Pick 2 digits x  s.t. $x \mod 3 = 1$: We must pick exactly 2 digits x s.t. $x \mod 3 = 2$.
Sum = $4! = 24$
Ans = $96 + 24 = 120$
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