<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] The two divisiblity rules you are assumed to use are: Divisible by $5$ if and only if last digit is $0$ or $5$ and; divisible by $3$ if and only if the sum of the digits add to a multiple of $3$.
Divisble by $5$ means last digit is $0$.  That's the only option.
Divisble by $3$ means that first four digits, lets call them $a,b,c,d$ add up to a multiple of $3$
We have already used $0$ so there we must pick from the remaining $6$ for digits that add to a multiple of $3$.
Insight is... we must remove $2$ and which $2$ can we remove?  Suppose we keep $a,b,c,d$ and remove $e,f$.  Then $a+b+c+d$ must be a multiple of $3$ and $(a+b+c+d) +(e+f) = 1+2+6 + 7 + 8 +9 = 33$. So we must have $e+f = 33 -$ a multiple of $3=$ a multiple of $3$.
So we can have $e$ and $f$ be $6$ and $9$ the only two multiples of $3$.  Or we can have $e$ or $f$ be a pair that aren't multiples of three but add to multiples of $3$.  That would mean one has remainder of $1$ and the other has a remainder of $2$. So one is $\{1,7\}$ then other is $\{2,8\}$.  So that is $4$ possible pairs.  Or it could be $\{e,f\}=\{6,9\}$ make $5$ possible pairs.
(If that was too confusing.  To have $e+f$ be a multiple of $3$ we can have $\{e,f\} = (1,2), (1,8), (7,2), (7,8), (6,9)$.  Five options. And $a,b,c,d$ can be...everything left.  Five options:  $(6,7,8,9), (2,6,7,9) ,(1,6,8,9),(1,2,6,9), (1,2,7,8)$.)
So there are $5$ choices for $(a,b,c,d)$ and there are $4! = 24$ ways to arrange them and there is only one option for the last digit.  There are $5*24 =120$ such numbers.
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