Show that any positive odd integer is of the form $6q+1$,$6q+3$,or $6q+5$,where $q$ is some integer.

(N/A) According to Euclid's Division Lemma,for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \leq r < b$.
Let $a$ be any positive odd integer and $b = 6$.
Substituting $b = 6$ in the lemma,we get $a = 6q + r$,where $0 \leq r < 6$.
This means the possible values for $r$ are $0, 1, 2, 3, 4, 5$.
If $r = 0$,then $a = 6q = 2(3q)$,which is even.
If $r = 1$,then $a = 6q + 1 = 2(3q) + 1$,which is odd.
If $r = 2$,then $a = 6q + 2 = 2(3q + 1)$,which is even.
If $r = 3$,then $a = 6q + 3 = 2(3q + 1) + 1$,which is odd.
If $r = 4$,then $a = 6q + 4 = 2(3q + 2)$,which is even.
If $r = 5$,then $a = 6q + 5 = 2(3q + 2) + 1$,which is odd.
Since $a$ is a positive odd integer,it cannot be of the form $6q, 6q+2,$ or $6q+4$. Thus,any positive odd integer must be of the form $6q+1, 6q+3,$ or $6q+5$.