Number of seven digit numbers without repetition of digits divisible by $3$ Find number of seven digit numbers divisible by $3$ with
$1.$ Repetition
$2.$ Without Repetition
For Part $1.$ The least seven digit number divisible by $3$ is $1000002$ and highest seven digit number is $9999999$
So total is $3000000$
For Part $2$ The least such number is $1023456$ and the next is $1023459$
Now sum of the digits of $1023456$ is $21$.
So if we include $7$ by removing one of the digits in $1023456$ then the digits that can be removed is $1$ or $4$.
If we include $8$ then digits that can be removed is $2$ or $5$.
but we get more cases here, any better way to approach?
 A: Let $N_7 = \text{what you are looking for}$
However, this consists of one set that doesn't use any zeros - $(NZ)_7$ - and another that does. The number in the set that uses zeros can be defined recursively as $6(NZ)_6$ because the zero can be placed in 6 different places.
Notice that $(NZ)_7 = \frac{\text{pair of two numbers that are divisible by 3}}{9c2} \cdot(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3)$
This is because the total sum of $1,2, ..8, 9$ is $45$ and we need to remove two numbers such that the sum of the others is still a multiple fo $3$ ie the two numbers we remove must be a multiple of $3$
Therefore, $(NZ)_7 = \frac{(NZ)_2}{9P2}(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3)$
Similarly, $(NZ)_6 = \frac{(NZ)_3}{9P3}(9\cdot8\cdot7\cdot6\cdot5\cdot4)$
$(NZ)_2$ can be counted, but also it can be expressed as $9\cdot3 - \frac{1}{9}\cdot 9 \cdot 3$. This is found because if the first is $0 \mod 3$, then there is an overcounting (1/3 of such cases are false $1/3 *1/3 = 1/9$)
Similarly, $(NZ)_2=9\cdot 8 \cdot 3 - \frac{2}{8} \cdot \frac{2}{3} \cdot 9 \cdot 8\cdot 3$. Here, the second part is found by considering the probability that the second number chosen is the same modulus as the first ( $2/8$), which results in only $1/3$ of the possible last digits (so we subtract out $2/3$)
Therefore, 
$$N_7 = \frac{9 \cdot 3 - 3}{72}(9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3) + 6 \cdot \frac{9\cdot8\cdot3-4\cdot9}{504}(9\cdot8\cdot7\cdot6\cdot5\cdot4) = 190080$$
This is definitely not my most eloquent answer, so please ask me questions if something is confusing, and I'll try to explain my thought process.
A: $(1)$ Number of possible digit combinations without $0$ digit: $\binom{9}{7}=36 $
$(2)$ Number of possible digit combinations including $0$ digit: $\binom{9}{6}=84 $
So there are $36 \cdot 7!+84 \cdot 6 \cdot 6!=544320$ numbers without digit repetition.
Now we have to find the numbers divisible by $3$.
We can seperate the cases:
$(1.a)$ $1$ digit divisible by $3$, $6$ digits not divisible by $3$: $\binom{3}{1}\binom{6}{6}=3\cdot1=3$ 
Out of the $\binom{6}{6}=1$ case 
$$(1,2,4,5,7,8)$$
$1$ is divisible by $3$, so there are  $\binom{3}{1}\cdot 1=3$  numbers divisible by $3$.
$(1.b)$ $2$ digits divisible by $3$, $5$ digits not divisible by $3$: $\binom{3}{2}\binom{6}{5}=3\cdot6=18$ 
Out of the $\binom{6}{5}=6$ cases
$$(1,2,4,5,7),(1,2,4,5,8),(1,2,4,7,8),(1,2,5,7,8),(1,4,5,7,8),(2,4,5,7,8)$$
$0$ is divisible by $3$, so there are  $\binom{3}{2}\cdot 0=0$ numbers divisible by $3$.
$(1.c)$ $3$ digits divisible by $3$, $4$ digits not divisible by $3$: $\binom{3}{3}\binom{6}{4}=1\cdot15=15$
Out of the $\binom{6}{4}=15$ cases
$$(1,2,4,5),...,(4,5,7,8)$$
$9$ is divisible by $3$, so there are  $\binom{3}{3}\cdot 9=9$ numbers divisible by $3$.
So in the $(1)$ case there are $(3+0+9)\cdot 7!=12\cdot 7!=60480$ numbers divisible by $3$.
Similarly we can find that in the $(2)$ case there are $=30\cdot 6\cdot 6!=129600$ numbers divisible by $3$.
Maybe it's not the best approach and for some numbers we have to test the 3-divisibility, but we got the answer: $60480+129600=190080$
