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

By Euclid’s division algorithm, b = a × q + r, 0 ≤ r < a

Here, b is any positive integer .

Let a be an arbitrary positive integer. Then corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that

a = 6q + r where 0 ≤ r < 6

Cubing both the sides using (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ a³ = (6q + r)³ = 216q³ + r³ + 3 × 6q × r(6q + r)

⇒ a³ = (216q³ + 108q²r + 18qr²) + r³

where 0 ≤ r < 6

Case I When r = 0 we get a³ = 216q³

⇒ a³ = 6(36q³) ⇒ a³ = 6m + 0 where m = 36q³ is an integer

Case II When r = 1 we get

a³ = (216q³ + 108q² + 18q) + 1

⇒ a³ = 6(36q³ + 18q² + 3q) + 1

⇒ a³ = 6m + 1, where m = (36q³ + 18q² + 3q) is an integer.

Case III When r = 2 we get

⇒ a³ = 6(36q³ + 36q² + 12q + 1) + 2

⇒ a³ = 6m + 2

where m = (36q³ + 36q² + 12q + 1) is an integer.

Case IV When r = 3 we get

a³ = (216q³ + 324q² + 162q) + 27

⇒ a³ = (216q³ + 324q² + 162q + 24) + 3

⇒ a³ = 6(36q³ + 54q² + 27q + 4) + 3

⇒ a³ = 6m + 3

where m = (36q² + 54q² + 27q + 4) is an integer.

Case V When r = 4 we get

a³ = (216q² + 432q² + 288q) + 64

⇒ a³ = 6(36q³ + 72q² + 48q) + 60 + 4

⇒ a³ = 6(36q³ + 72q² + 48q + 10) + 4

⇒ a³ = 6m + 4

where m = (36q³ + 72q² + 48q + 10) is an integer.

Case VI When r = 5 we get

a³ = (216q³ + 540q² + 450q) + 120 + 5

⇒ a³ = 6(36q³ + 90q² + 75q + 20) + 5

⇒ a³ = 6m + 5

where m = (36q³ + 90q² + 75q + 20) + 5

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 forms 6m, 6m + 1, 6m + 2, 6m + 3,6m + 4 and 6m + 5i.e., 6m + r.