Question

show that the Cube of a positive integer
is of the form 6q+r where q is
an integer and r = 0, 1,2,3,4,5​

Answer 1

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) = 216 q3 = 6(36q) 3 + 0

= 6m + 0, where m is an integer.

(6q+1)3 = 216 q3 + 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, 5is also of the form 6m + r.

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