Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let ‘a’ be any positive integer and b = 6.
∴ By Euclid’s division algorithm, we have
a = bq + r, 0 ≤ r ≤ b
a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4,5
Now, ‘a’ may be of the form of 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5
If ‘a’ is of the form 6q, 6q + 2, 6q + 4 then ‘a’ is an even.
In above we can see clearly that the numbers of the form 6q, 6q + 2, 6q + 4 are having the factor 2.
∴ The numbers of the form 6q, 6q + 2, 6q + 4 are even.
If ‘a’ is of the form 6q +1,6q +3, 6q + 5 then ‘a’ is an odd.
As if
∵ We know that the number of the form 2k + 1 is odd.
∴ The numbers of the form 6q + 1, 6q + 3, 6q + 5 are odd.