Solution:

Let’s say that $n$ be any arbitrary positive odd integer.
Upon dividing $n$ by 6, let $m$ be the quotient and $r$ be the remainder.
Therefore, by Euclid’s division lemma, we get
$n=6 m+r$, where $0 \leq r<6$
As $0 \leq r<6$ and $r$ is an integer, $r$ can take values $0,1,2,3,4,5$
$\Rightarrow n=6 m$ or $n=6 m+1$ or $\mathrm{n}=6 m+2$ or $n=6 m+3$ or $n=6 m+4$ or $n=6 m+5$
But $n \neq 6 m$ or $n \neq 6 m+2$ or $n \neq 6 m+4(\because 6 m, 6 m+2,6 m+4$ are multiples of 2, so an even integer whereas $n$ is an odd integer)
$\Rightarrow n=6 m+1 \text { or } n=6 m+3 \text { or } n=6 m+5$
As a result, any positive odd integer is of the form $(6 m+1)$ or $(6 m+3)$ or $(6 m+5)$, where $m$ is some integer.