Let ‘a’ be any positive integer.
*Then from Euclid’ division lemma,**
a = bq+r ; where 0 < r < b
Putting b=6 we get,
⇒ a = 6q + r, 0 < r < 6
For r = 0, we get a = 6q = 2(3q) = 2m, which is an even number. [m = 3q]
For r = 1, we get a = 6q + 1 = 2(3q) + 1 = 2m + 1, which is an odd number. [m = 3q]
For r = 2, we get a = 6q + 2 = 2(3q + 1) = 2m, which is an even number. [m = 3q + 1]
For r = 3, we get a = 6q + 3 = 2(3q + 1) + 1 = 2m + 1, which is an odd number. [m = 3q + 1]
For r = 4, we get a = 6q + 4 = 2(3q + 2) + 1 = 2m + 1, which is an even number. [m = 3q + 2]
For r = 5, we get a = 6q + 5 = 2(3q + 2) + 1 = 2m + 1, which is an odd number. [m = 3q + 2]
Thus, from the above it can be seen that any positive odd integer can be of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.
