Question

Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5,
where q is some integer.​

Answer 1

Answer:

a= bq + r

b = 6

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

eq - 1

a = bq + r

=6q + 0

( even)

eq - 2

a = bq + r

= 6q + 1

( odd )

eq. - 3

a. = bq + r

= 6q + 2

=2(3q + 1)

( even )

eq - 4

a = bq + r

=6q + 3

= 3 (2q + 1)

( even)

Answer 2

$\huge \underline \mathbb {SOLUTION:-}$

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.