How many five-digit numbers can be formed from the digits 0, 1, 2, 3, 4, and 5, if no digits can repeat and the number must be divisible by 4?

a.36
b.48
c.72
d.96
e.144

OA is E

I could conclude the the last 2 digits of the 5-digit nos. should be 04 or 12 or 20 or 24 or 32 or 40 or 52 in order for them to be divisible by 4.

Now suppose the last 2 digits are 04,20 or 40

_ _ _ _ _--> the last 2 positions can be filled in 3 ways. The first 3 however can be filled in 4C3 ways giving us 4C3*3= 12 nos.

For the remaining combinations i.e. having last 2 digits 12,24,32 or 52:

_ _ _ _ _ --> the last 2 positions can be filled in 4 ways. The first position cannot be 0. Therefore we can fill the first position in 3 ways. The remaining 2 (digit 2nd &3rd) can be filled in 3C2 ways giving us 3*3C2*4= 36 ways.

Total 36+12= 48 nos. Obviously this is wrong. Please guide.