Question

how to do 2nd question

Answer 1

By Euclid division lemma

a=bq+r where 0<r<b

so a=6q+r where o<r<6

then r=0,1,2,3,4,5

if r=0 then a=6q+0

=6q

if r=1 then a=6q+1

if r=2 then a=6q+2

=2(3q+1)

if r=3 then a=6q+3

=3(2q+1)

if r=4 then a=6q+4

=2(3q+2)

if r=5 then a=6q+5

Hence , any positive odd integer is of the form 6q+1,or 6q+3,or6q+5,where q is some integer