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

Let a be an arbitrary positive integer.

Then, by Euclid’s division algorithm, 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

`\implies` a³ = (6q + r)³ = 216q³ + r³ + 3 . 6q . r(6q + r)  .......[∵ (a + b)³ = a³ + b³ + 3ab(a + b)]

`\implies` a³ = (216q³ + 108q²r + 18qr²) + r³   .......(i)

Where 0 < r < 6

Case I: When r = 0,

Then putting r = 0 in equation (i), we get

a³ = 216q³ 

= 6(36q³)

= 6m

Where, m = 36q³ is an integer.

Case II: When r = 1,

Then putting r = 1 in equation (i), we get

a³ = (216q³ + 108q² + 18q) + 1

= 6(36q³ + 18q² + 3q) + 1

`\implies` a³ = 6m + 1

Where, m = (36q³ + 18q² + 3q) is an integer.

Case III: When r = 2,

Then putting r = 2 in equation (i), we get

a³ = (216q³ + 216q² + 72q) + 8

a³ = (216q³ + 216q² + 72q + 6) + 2

`\implies` a³ = 6(36q³ + 36q² + 12q + 1) + 2

= 6m + 2

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

Case IV: When r = 3,

Then putting r = 3 in equation (i), we get

a³ = (216q³ + 324q² + 162q) + 27

= (216q³ + 324q² + 162q + 24) + 3

= 6(36q³ + 54q² + 27q + 4) + 3

= 6m + 3

Where, m = (36q³ + 54q² + 27q + 4) is an integer.

Case V: When r = 4,

Then putting r = 4 in equation (i), we get

a³ = (216q³ + 432q² + 288q) + 64

= 6(36q³ + 72q² + 48q) + 60 + 4

= 6(36q³ + 72q² + 48q + 10) + 4

= 6m + 4

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

Case VI: When r = 5,

Then putting r = 5 in equation (i), we get

a³ = (216q³ + 540q² + 450q) + 125

`\implies` a³ = (216 q³ + 540q² + 450q) + 120 + 5

`\implies` a³ = 6(36q³ + 90q² + 75q + 20) + 5

`\implies` a³ = 6m + 5
Where, m = (36q³ + 90q² + 75q + 20) 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, 6m + 1, 6m + 2, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.