Show that any odd positive integer …

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r < b. Let a be the positive odd integer which when divided by 6 gives q as quotient and r as remainder. According to Euclid’s division lemma a = bq + r a = 6q + r………………….(1) where, (0 ≤ r < 6) So r can be either 0, 1, 2, 3, 4 and 5. Case 1: If r = 1, then equation (1) becomes a = 6q + 1 The Above equation will be always as an odd integer. Case 2:  If r = 3, then equation (1) becomes a = 6q + 3 The Above equation will be always as an odd integer. Case 3:  If r = 5, then equation (1) becomes a = 6q + 5 The above equation will be always as an odd integer. ∴ Any odd integer is of the form  6q + 1 or 6q + 3 or 6q + 5. Hence proved.