Question

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 1

Solution:
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 of each term, we have,
(6q)³ = 216 q³ = \(6\times36q^3\) + 0
= 6m + 0, where m is an integer
(6q+1)³ = 216q³ + 108q² + 18q + 1
= 6(36q³ + 18q² + 3q) + 1
= 6m + 1, where m is an integer
(6q+2)³ = 216q³ + 216q² + 72q + 8
= 6(36q³ + 36q² + 12q + 1) +2
= 6m + 2, where m is an integer
(6q+3)³ = 216q³ + 324q² + 162q + 27
= 6(36q³ + 54q² + 27q + 4) + 3
= 6m + 3, where m is an integer
(6q+4)³ = 216q³ + 432q² + 288q + 64
= 6(36q³ + 72q² + 48q + 10) + 4
= 6m + 4, where m is an integer
(6q+5)³ = 216q³ + 540q² + 450q + 125
= 6(36q³ + 90q² + 75q + 20) + 5
= 6m + 5, where m is an integer
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.