Show that every positive even integer is of the form (6m+1) or (6m+3) or (6m+5)where m is some integer.
Show that every positive even integer is of the form (6m+1) or (6m+3) or (6m+5)where m is some integer.

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.