How many $3$-digit numbers are divisible by $4$? 
I want to find out how many numbers can be formed taking the digits from zero to nine without repetition such that they are divisible by four (of 3 digits).

My Work: I find out the number pairs for the ones and ten's places which are divisible by four (like $04, 40, 08, 80, 12, 16,$ etc). These are $22$ in number. For any pair, for the hundred place, I have $8$ digits left (including $0$). So, my rough calculation becomes $8\times 22 = 176$. Then, I find out the number pairs without zero ($16$ in number)  . 
Thus, my final answer becomes $176 - 16\times 1 = 160$. Am I correct ?
 A: 
How many three-digit positive integers with distinct digits are divisible by $4$?

If this was the intended question, then you are indeed correct.  Your explanation could be improved if you made clear that you were subtract the $16$ strings with leading digit zero. 
For a number to be divisible by $4$, the number formed by its last two digits (appending leading zeros as necessary) must be divisible by $4$.  
\begin{array}{c c c c c}
\color{red}{00} & \color{blue}{04} & \color{blue}{08} & 12 & 16\\
\color{blue}{20} & 24 & 28 & 32 & 36\\
\color{blue}{40} & \color{red}{44} & 48 & 52 & 56\\
\color{blue}{60} & 64 & 68 & 72 & 76\\
\color{blue}{80} & 84 & \color{red}{88} & 92 & 96
\end{array}
Of the $25$ such numbers, three of them have a repeated digit ($\color{red}{00}, \color{red}{44}, \color{red}{88}$).  Of the remaining $22$ choices, exactly six include the digit $0$ ($\color{blue}{04}, \color{blue}{08}, \color{blue}{20}, \color{blue}{40}, \color{blue}{60}, \color{blue}{80}$).
Since the leading digit of a three-digit positive integer cannot be $0$, in the absence of the restriction that the digits cannot be repeated, we would have nine choices for the leading digit.  Given that restriction, we have $9 - 1 = 8$ choices for the leading digit when $0$ is one of the final two digits and $9 - 2 = 7$ choices for the leading digit when $0$ is not one of the final two digits.  Since there are six admissible choices for the final two digits that include the digit $0$ and $22 - 6 = 16$ admissible choices for the final two digits that do not include the digit $0$, the number of three-digit numbers that are divisible by $4$ in which no digit is repeated is 
$$8 \cdot 6 + 7 \cdot 16 = 48 + 112 = 160$$
A: HINT: 
There are $\lfloor2.5-\epsilon\rfloor=2$ single digit multiples of four. (excluding $0$)
There are $\lfloor25-\epsilon\rfloor-2=22$ two digit multiples of four.
There are $\lfloor 250-\epsilon\rfloor-22-2=225$ three digit multiples of four.
You now have to subtract any cases where there are repeated digits.
