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 $3$?

 $120$

 A: 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 $5\times 4! = 120$
A: $0$ 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 = $96 + 24 = 120$
A: 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.
A: 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 $3$.
