Question

use Euclid division Lemma to show that any positive odd integer is of the Form 6q + 1 or 6q+ 3 or6q + 5 Where Q is some integer

Answer 1

⭐⭐⭐⭐⭐Hi mate your answer is.... ⭐⭐
${} \huge{answer}$
Let a be any arbitrary odd positive integer and
b=6.
Then, a=bq+r
a=6q+r,
q is the whole number and r is any integer
such that 0<r<6
Now, the possible values of r is 0,1,2,3,4,5
If r=0 then a=6q
If r=1 then a=6q+1
If r=2 then a=6q+2
If r=3 then a=6q+3
If r=4 then a=6q+4
If r=5 then a=6q+5
But 6q, 6q+2, and 6q+4 are even integers
and a is an odd integers. Thus, a can't be form
6q, 6q+2 or 6q+4.
Hence, a=6q+1, 6q+3 or 6q+5 for some integer
q.
${} \red{i \: hope \: it \: will \: help \: you}$
⭐⭐⭐⭐⭐⭐⭐⭐⭐