Question

Show that cube of any positive integer
of form 6q+r q is an integer and r 0,1,2,3,4,5 is also in form of 6m+r

Answer 1

firstly let a be any positive Integer
and b= 6
so r=0,1,2,3,4,5,
so by Euclid division algorithm
a=bq+r
a=6q+0=6q
a=6q+1
a=6q+2,6q+3,6q+4,6q+5
Taking cube of each term, we have,
(6q)3 = 216 q3 = 6(36q)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.