How many 4 digit number can be formed by 0,1,2,3,4,5 divisible by 4 
How many 4 digit number can be formed by 0,1,2,3,4,5 divisible by 4 with repetition

My Approach:
Last two digits can be 00,04,12,20,24,32,40,44,52 that is 9 possibilities for last two digits.
For the hundredth place digit all 6 possibilities exist.For the thousand place we have 5 options (1,2,3,4,5 ). 
Hence the solution is 5*6*9=270
Is my solution correct? Please correct me if I am wrong.
 A: Your ans of $270$ is correct, but you might find this way simpler:
Since a number can't start with zero, there will be $5\cdot6^3$ valid numbers,
of which exactly $1/4$ will be divisible by $4$
[Think of a number system to base 6 to understand the rationale]
A: There are 5*6*6*6 = 1080 possible numbers.
Half of which, 540, is divisible by 2.
Half of that again, 270, is divisible by 4.
Edit: Perhaps not obvious. But we can observe, that the numbers are consecutive, except gaps of length 4, 44, or 444; all gaps multiple of 4.
Edit2: 
In a sequence of consecutive numbers every fourth number is divisible by 4.
Modulo 4 the sequence goes 0, 1, 2, 3, 0, 1 ...
In the increasing sequence of numbers made from only the digits 0, 1, 2, 3, 4 and 5, there will be gaps. We can observe that the gaps will have lengths of 4, 44 or 444. All gaps are a multiple of 4.
Therefore the modulo 4 sequence must run systematically and always increase by 1 modulo 4 across a gap.
There are 5*6*6*6 = 1080 four digit numbers in the sequence. A multiple of 4, so 1/4, i.e. 270 of the numbers, are divisible by 4.
A: Best way is to identify the first(1000) and last 4 digit numbers(5552) divisible by 4. Then they will be in arithmetic progression with d=4. Then you can identify n=number of terms[(5552-1000)/4+1=1139 numbers can be formed], which gives you exact answer for any such problems with repetition.
