Question

Show that any positive odd integer is in the form of

6q + 1,6q + 3 or 6q + 5 where q is any integer.​

Answer 1

Answer:

Let a be any real no.

b=6

By Euclids Division Lemma

a=bq+r

=0<r<b

The possible values of r are 1 2 3 4 and 5

a=6q+1 when r=1

By using this technique we can get the other number

Answer 2

$\large\underline{\sf{Solution-}}$

By Euclid Lemma,

Let there exist integers q and r such that a = 6q + r, where r = 0, 1, 2, 3, 4, 5.

So, 6 cases arises.

Here, below the 6 cases,

a = 6q = 2(3q), so its even.

a = 6q + 1 = 2(3q) + 1, so its odd.

a = 6q + 2 = 2(3q + 1), so its even.

a = 6q + 3= 2(3q) + 2 + 1 = 2(3q + 1) + 1, so its odd

a = 6q + 4= 2(3q + 2) + 1, so its even

a = 6q + 5 = 2(3q) + 4 + 1 = 2(3q + 2) + 1, so its odd.

Hence, we concluded that any positive odd integer is of the form 6q + 1, 6q + 3, 6q + 5.