Show that any positive odd integer is of the form of 6q+1,or6q+3,or 6q+5,where q is some integer
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Let a be any positive integer and b = 6∴ by Euclid’s division lemmaa = bq + r, 0≤ r and q be any integer q ≥ 0∴ a = 6q + r,where, r = 0, 1, 2, 3, 4, 5If a is even then then remainder by division of 6 is 0,2 or 4Hence r=0,2,or 4or A is of form 6q,6q+2,6q+4As, a = 6q = 2(3q), ora = 6q + 2 = 2(3q + 1), ora = 6q + 4 = 2(3q + 2).If these 3 cases a is an even integer.but if the remainder is 1,3 or 5 then r=1,3 or 5or A is of form 6q+1,,6q+3 or,6q+5Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,\xa0Case 2: a = 6q + 3 = 6q + 2 + 1,= 2(3q + 1) + 1 = 2n + 1,\xa0Case 3: a = 6q + 5 = 6q + 4 + 1= 2(3q + 2) + 1 = 2n + 1This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.