Prove that any positive odd integer is of the form $6m+1$ or $6m+3$ or $6m+5$,where $m \in N \cup \{0\}$.

(N/A) Let $a$ be any positive odd integer. By Euclid's division lemma,for any positive integer $a$ and $b=6$,there exist unique integers $q$ and $r$ such that $a = 6q + r$,where $0 \le r < 6$.
Since $r$ can be $0, 1, 2, 3, 4, 5$,the possible forms of $a$ are $6q, 6q+1, 6q+2, 6q+3, 6q+4,$ and $6q+5$.
If $a = 6q, 6q+2,$ or $6q+4$,then $a$ is even because these are divisible by $2$ (e.g.,$6q = 2(3q)$).
Since $a$ is an odd integer,it cannot be of the form $6q, 6q+2,$ or $6q+4$.
Therefore,any positive odd integer must be of the form $6q+1, 6q+3,$ or $6q+5$ for some integer $q \ge 0$. Replacing $q$ with $m$,we get the forms $6m+1, 6m+3,$ or $6m+5$.