Show that any positive odd integer is of the form

Solution:

Let us start with taking a, where a is any positive odd integer. We apply the divisionalgorithm, with a and b = 6. Since $0 \leq \mathrm{r}<6$, the possible remainders are 0, 1, 2, 3, 4, 5. That is, a can be 6q or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is the quotient. However, since a is odd, we do not consider the cases 6q, 6q + 2 and 2 Real Number 6q + 4 (since all the three are divisible by 2). Therefore, any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5.