<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] To be divisible by $5$, the five-digit number must end in $0$. There are $360=6\cdot5\cdot4\cdot3$ integers that can be made by arranging four of the available non-zero digits before the zero. These $360$ five-digit numbers might have remainder $0$, $1$, or $2$ when divided by $3$, and since the six available digits are equally distributed modulo $3$ (two of each possibility mod $3$), each of the three possible remainders is equally likely among the $360$ possible five-digit numbers. Therefore exactly one third of them, or $120$, are divisible by $3$.
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