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. Alternatively: Let a be any odd positive integer we need to prove that a is of the form 6q + 1 , or 6q + 3 , or 6q + 5 , where q is some integer. Since a is an integer consider b = 6 another integer applying Euclid’s division lemma we get a = 6q + r for some integer q ≤ 0, and r = 0, 1, 2, 3, 4, 5 since 0 ≤ r < 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4 (since all these are divisible by 2) Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, where k1 is a positive integer 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers. Therefore, any odd integer can be expressed is of the form 6q + 1, or 6q + 3, or 6q + 5 where q is some integer