Question

Use division algorithm to show that any positive odd integer is of the form 69+
or 69 + 3 or 69 + 5, where q is some integer.​

Answer 1

$\bigstar$ 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.

Answer 2

Let

′

a

′

be any positive integer and b=6

Then by division algorithm

a=6q+r where r=0,1,2,3,4,5

so, a is of the form 6q or 6q+1 or 6q+2 or 6q+3 or

6q+3 or 6q+4 or 6q+5

Therefore If s is an odd integer

Then

′

a

′

is of the form 6q+1 or 6q+3 6q+5

Hence a positive odd integer is of the form 6q+1 or 6q+3 or 6q+5