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

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.