Divisibility of numbers without a digit 
How many of the integers from $0,1, 2, ... ,999$ are neither divisible by $9$ nor contain the digit $9$. 

Let $N$ be an integer, so, $N \equiv 1, 2, 3, 4, 5, 6, 7, 8 \pmod{9}$. 
That is $8$ numbers in a cycle of $9$. From $1 \to 999$ there are: $11$ cycles of $9$ so, there are $8*111 = 888$ numbers not divisible by $9$ also counting the $0$ from $0 \to 999$ there are $889$ not divisible by $9$. 
But this also includes the ones containing a $9$ eg. $39$. How to get about this?
Hint only please!
 A: Let $A$ denote the set of integers from $0$ to $999$ which contain a $9$
Let $B$ denote the set of integers from $0$ to $999$ which are divisible by $9$
The total you are seeking is:
$|A^c\cap B^c|$
By properties of complements and inclusion-exclusion this simplifies:
$|A^c\cap B^c| = |\Omega| - |A\cup B| = |\Omega| - |A|-|B|+|A\cap B|$
where $\Omega$ is the set of numbers $\{0,1,\dots,999\}$
You should be capable of counting $|\Omega|, |A|, |B|$ with little difficulty.  What remains is to calculate $|A\cap B|$
To do this, I recommend breaking into cases as to the number of $9$'s present and further break into cases based on what the first digit is.  Remember that a number divisible by $9$ must have the digitsum be a multiple of $9$.

More details:
Calculating $|\Omega|$

 It is trivially equal to $1000$

Calculating $|A|$

 Instead of counting how many contain a $9$, we count how many don't.  There are $9$ digits to choose from $\{0,1,2,\dots,8\}$.  By multiplication principle, $9^3=729$ numbers do not contain a $9$, implying that $1000-729 = 271$ numbers do contain a $9$

Calculating $|B|$

 There are $1+\lfloor\frac{999}{n}\rfloor$ numbers divisible by $n$ in the range $0\leq n\leq 999$, so in the case of divisibility by $9$, we have $|B|=112$

Calculating $|A\cap B|$

 Break into cases: Exactly $3$ nines.  There is one such possibility (999).  Exactly $2$ nines.  This is only possible if the remaining digit is a zero.  There are then three such possibilities (99, 909, 990).  One digit is a 9.  Pick the location of the nine (3 positions).  Pick the earliest non-9 digit.  Suppose the digit picked is $x$.  Regardless which is picked, there is only one choice for the final remaining digit (namely $0$ in the case that $x=0$ and $9-x$ in all other cases).  There are then a total of $3\cdot 9=27$ possibilities in this case, for a combined total of $|A\cap B|=1+3+27 = 31$

Giving a final answer of:

 $1000-271-112+31 = 648$

A: Note that the numbers that doesn't contain the digit $9$ are :$8+72+648=728$ Write the number $n$ in basis $10$: $n=10^2a_2+10a_1+a_0$ and consider the number in $mod 9$. If the number is of $3-digits$ the sum of $a_2+a_1+a_0$ doesn't have to be equal to $9$ and $18$. To choose the first digit($a_2$) you have $8$  possible ways of choice. For the second digit you have $9$ possible ways of choice. While for the third digit you have $7$ possible ways of choice if the sum of the first and second digit isn't equal to $9$. Indeed exist only a number  from $1$ to $8$ such that $n$ is then divisible for $9$. For example if $a_2=5$ and $a_1 =3$ (note that $a_2+a_1$ isn't equal to $9$) then exist only a number from $1$ to $8$ and is $1$ such that the number is divisible for $9$. If the sum of first and second I digit is equal to $9$ then you have $8$ possible ways of choice. A similar argument you can use for a number of $2$-digit. Now you have to use a bit of combinatorics.
