Let ‘a’ be an odd positive integer.

Let us now apply division algorithm with a and b = 6.

∵ 0 ≤ r < 6, the possible remainders are 0, 1, 2, 3, 4 and 5.

i.e., ’a’ can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5, where q is the quotient.

But ‘a’ is taken as an odd number.

∴ a can’t be 6q or 6q + 2 or 6q + 4.

∴ Any odd integer is of the form 6q + 1, 6q + 3 or 6q + 5.