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Show that any odd positive integer in the form of 6q+1, 6q+3, or 6q+5 |
| Answer» According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers\xa0q\xa0and\xa0r\xa0which satisfies the condition\xa0a = bq + r\xa0where 0\xa0≤ 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 lemmaa = bq + ra = 6q + r………………….(1)where, (0\xa0≤ r\xa0< 6)So r can be either 0, 1, 2, 3, 4 and 5.Case 1:If r = 1, then equation (1) becomesa = 6q + 1The Above equation will be always as an odd integer.Case 2:\xa0If r = 3, then equation (1) becomesa = 6q + 3The Above equation will be always as an odd integer.Case 3:\xa0If r = 5, then equation (1) becomesa = 6q + 5The 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. | |