Question

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

Answer 1

If a and b are two positive integers such that a is greater than b; then according to Euclid's division algorithm; we have

a = bq + r; where q and r are positive integers and 0 ≤ r < b.

Let b = 6, then

a = bq + r ⇒ a = 6q + r; where 0 ≤ r < 6.

When r = 0 ⇒ a = 6q + 0 = 6q; which is even integer

When r = 1 ⇒ a = 6q + 1  which is odd integer

When r = 2 ⇒ a = 6q + 2  which is even.

When r = 3 ⇒ a = 6q + 3  which is odd.

When r = 4 ⇒ a = 6q + 4  which is even.

When r = 5 ⇒ a = 6q + 5  which is odd.

This verifies that when r = 1 or 3 or 5; the integer obtained is 6q + 1 or 6q + 3 or 6q + 5 and each of these integers is a positive odd number.

Hence the required result.