The correct options are
A 6q
B 6q+2
D 6q+4
Let 'a' be any positive integer and b = 6.
Then by Euclid's algorithm,
a = 6q + r, for some integer q >= 0 and
0 <= r <= 5
Then a can be,
6q , 6q + 1, 6q + 2, 6q + 3,
6q + 4, 6q + 5.
Since 'a' is even, 'a' cannot be 6q + 1, 6q + 3, 6q + 5 (since these are not divisible by 2).
Therefore, any positive even integer is of the form 6q, 6q + 2 or 6q + 4.