Correct Answer - C
We cannnot write 903.
If in the selection of n cards, we get either
(9 or 3), (9 or 0), (0 or 3), (only 0), (only 3) or (only 9).
For (9 or 3) can be selected `=2xx2xx2xx . . .xxn` factors `=2^(n)`
Similarly, (9 or 0) or (0 or 3) can be selected `=2^(n)`
In the above selection (only 0) or (only 3) or (only 9) is repeated twice.
`therefore`Total ways `=2^(n)+2^(n)+2^(n)-3=93`
`implies3*2^(n)=96implies2^(n)=32=2^(5)`
`thereforen=5`.