Let a be any positive integer
Let b = 6
By dividing algorithm rule for a and b : a = 6q + r, for q > 0,and 0 ≤ r <
The possible remainders are 0, 1, , , ,
So, a = 6q , 6q+1, 6q+2, 6q+3, where q is the
However, since a is
a cannot be 6q = 2(3q) or 6q+2 = 2(3q+1), since they were divisible by
Therefore, any odd integer is of the form 6q + or 6q + or 6q +