{"id":10343,"date":"2022-04-01T17:28:57","date_gmt":"2022-04-01T11:58:57","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10343"},"modified":"2022-04-05T18:24:05","modified_gmt":"2022-04-05T12:54:05","slug":"show-that-any-positive-odd-integer-is-of-the-form-6q1-or-6q3-or-6q5-where-q-is-some-integer","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/show-that-any-positive-odd-integer-is-of-the-form-6q1-or-6q3-or-6q5-where-q-is-some-integer\/","title":{"rendered":"Show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5, where q is some integer."},"content":{"rendered":"
By Euclid’s division algorithm, we have<\/p>\n
a = bq + r\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(i)<\/p>\n
On putting, b = 6 in (1), we get<\/p>\n
a = 6q + r\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[0 \\(\\le\\) r < 6]<\/p>\n
If r = 0, a = 6q, 6q is divisible by 6\u00a0 \\(\\implies\\)\u00a0 6q is even.<\/p>\n
If r = 1, a = 6q + 1, 6q + 1 is not divisible by 2.<\/p>\n
If r = 2, a = 6q + 2, 6q + 2 is divisible by 2\u00a0 \\(\\implies\\)\u00a0 6q + 2 is even.<\/p>\n
If r = 3, a = 6q + 3, 6q + 3 is not divisible by 2.<\/p>\n
If r = 4, a = 6q + 4, 6q + 4 is divisible by 2\u00a0 \\(\\implies\\)\u00a0 6q + 4 is even.<\/p>\n
If r = 5, a = 6q + 5, 6q + 5 is not divisible by 2.<\/p>\n
Since, 6q, 6q + 2, 6q + 4 are even.<\/p>\n
Hence, the remaining integers are 6q + 1, 6q + 3, 6q + 5 are odd.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : By Euclid’s division algorithm, we have a = bq + r\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0……….(i) On putting, b = 6 in (1), we get a = 6q + r\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[0 \\(\\le\\) r < 6] If r = 0, a = 6q, 6q is …<\/p>\n