Question

Prove that the cube of a positive integer of the form 6m+r, where m is an integer and r= 0, 1, 2, 3, 4, 5 is also of the form 6q+r.

Answer 1

Case (v) : When r = 5,
(6m + r) ³= (6m + 5)³
= (6m) ³ + 3 × 6m × 5(6m + 5) + (5)³
= (6m) ³ + 3 × 6m × 5(6m + 5) + 120 + 5
= 6[36m³ + 15m(6m + 5) + 20] + 5
= 6q + 5, where q = 36m³ + 15m(6m + 5) +20

Case (iv) : When r = 0,
(6m + r) ³ = 216m³
= 6(36m³)
= 6q + 0, where q = 36m³

By all the six cases it is proved that the cube of a positive integer of the form 6m + r, where m is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6q + r.

Hope this helps you.....:)