Question

Show that any positive odd integer is of the form ( 6m + 1 ) or ( 6m + 3 ) or ( 6m + 5 ) where m is some integer

Answer 1

Multiplying any number with 2 or it's multiple is always an even number and adding an odd number to even number is always odd.

Answer 2

HOLA !!

===============
===============

Let m be a arbitrary integer

Where m divide 6 , to get remainder as r and to get qoutient as q

➡ By Euclid's lemma ( A = bq + r )

( 6m + 1 ) .. ( 6m + 3 ) .. ( 6m + 5)

➡ ( 6m + 1 ) = 7 .. were 7 is odd

➡ ( 6m + 3 ) = 9 .. were 9 is odd

➡ ( 6m + 5 ) = 11 .. were 11 is odd

Hence , any postive odd integer can be of the form ( 6m + 1 ) .. ( 6m + 3).. ( 6m + 5)

===================
===================

HOPE U UNDERSTAND ☺☺☺