Question

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

Answer 1

Answer:

let a be any odd positive integers and b =6

it means

$a = 6q + r$

by algorithm

$a = 6q + r \: (1)e {e}^{2}$

and we know we find the remainder

: so the possible remainder are 0,1,2,3,4,5(2) becuz the remainders are always less than divisor and ultimately more than 0

Step-by-step explanation:

1 &2

if r=0 so the a is 6q =:a=6q

when r = 1 then a = 6q+1

when r = 2 then a= 6q+2

when r =3 then a =6q+3

when r =4 then a = 6q+4

when r = 5 then a = 6q+5

$so \: the \: even \: integars \: are \: \\ a = 6q \\ a = 6q + 2 \\ a = 6q + 4$

hence when a is odd it is in the form of 6q+1 ,6q+3,6q+5 for some integar

;hence proved :

Answer 2

Answer:

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.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly,

6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or

6q + 3, or 6q + 5