

InterviewSolution
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Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r. |
Answer» Given, 6q + r is a positive integer, where q is an integer and r = 0, 1, 2, 3, 4, 5 Then, the positive integers are of the form 6q, 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5. Taking cube on L.H.S and R.H.S, For 6q, (6q)3 = 216 q3 = 6(36q)3 + 0 = 6m + 0, (where m is an integer = (36q)3) For 6q+1, (6q+1)3 = 216q3 + 108q2 + 18q + 1 = 6(36q3 + 18q2 + 3q) + 1 = 6m + 1, (where m is an integer = 36q3 + 18q2 + 3q) For 6q+2, (6q+2)3 = 216q3 + 216q2 + 72q + 8 = 6(36q3 + 36q2 + 12q + 1) +2 = 6m + 2, (where m is an integer = 36q3 + 36q2 + 12q + 1) For 6q+3, (6q+3)3 = 216q3 + 324q2 + 162q + 27 = 6(36q3 + 54q2 + 27q + 4) + 3 = 6m + 3, (where m is an integer = 36q3 + 54q2 + 27q + 4) For 6q+4, (6q+4)3 = 216q3 + 432q2 + 288q + 64 = 6(36q3 + 72q2 + 48q + 10) + 4 = 6m + 4, (where m is an integer = 36q3 + 72q2 + 48q + 10) For 6q+5, (6q+5)3 = 216q3 + 540q2 + 450q + 125 = 6(36q3 + 90q2 + 75q + 20) + 5 = 6m + 5, (where m is an integer = 36q3 + 90q2 + 75q + 20) Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r. |
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