Let's take 'a' as any positive integer and b = 6.
Then using Euclid’s algorithm we get a = 6q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < b and the value of b is 6.
So total possible forms will be 6q + 0 , 6q + 1 , 6q + 2,6q + 3, 6q + 4, 6q + 5
6q + 0
6 is divisible by 2 so it is an even number.
6q + 1
6 is divisible by 2 but 1 is not divisible by 2 so it is an odd number.
6q + 2
6 is divisible by 2 and 2 is also divisible by 2 so it is an even number.
6q +3
6 is divisible by 2 but 3 is not divisible by 2 so it is an odd number .
6q + 4
6 is divisible by 2 and 4 is also divisible by 2 it is an even number.
6q + 5
6 is divisible by 2 but 5 is not divisible by 2 so it is an odd number.
So, any positive odd integer will be of the form 6q + 1, or 6q + 3, or 6q + 5.