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.