By taking,’ a’ as any positive integer and b = 6.

Applying Euclid’s algorithm

a = 6q + r

Here,  r = remainder = 0, 1, 2, 3, 4, 5 and q ≥ 0

So, total possible forms are  6q + 0, 6q + 1, 6q + 2, 6q + 3, 6q + 4 and 6q + 5

6q + 0 , (6 is divisible by 2, its an even number)

6q + 1, ( 6 is divisible by 2 but 1 is not divisible by 2, its an odd number)

6q + 2, (6 and 2 both are divisible by 2, its an even number)

6q + 3, (6 is divisible by 2 but 3 is not divisible by 2, its an odd number)

6q + 4, ( 6 and 4 both are divisible by 2, its an even number)

6q + 5 , (6 is divisible by 2 but 5 is not divisible by 2, its an odd number)

Therefore, odd numbers will be in the form 6q + 1, or 6q + 3, or 6q+5