Question

use euclids division lemma to show that the cube of an any positive integer is either of the 9m,9m+1,9m+8 for some integer m​

Answer 1

Question: Use Euclid's Division Lemma to show that the cube of an any positive integer is either of the 9m, 9m+1, 9m+8 for some integer m​.

By using Euclid's Division Lemma,

a = bq + r where 0 ≤ r < b

Let b = 3,

⇒ a = 3q + r where 0 ≤ r < b

And r = 0, 1, 2

Case 1

When r = 0,

a = bq + r

a = 3q

Cubing on both sides we get,

a³ = (3q)³

a³ = 9q³

Where m = q³

a³ = 9m → 1

Case 2

When r = 1,

a = bq + r

a = 3q + 1

Cubing on both sides we get,

a³ = (3q + 1)³

Using (a + b)³ = a³ + b³ + 3ab(a + b) we get,

a³ = (3q)³ + (1)³ + 3(3q)(1)[3q + 1]

a³ = 27q³ + 1 + 9q[3q + 1]

a³ = 27q³ + 1 + 27q + 9q

a³ = 27q³ + 27q + 9q + 1

a³ = 9(3q³ + 3q + q) + 1

If m = 3q³ + 3q + q

a³ = 9m + 1 → 2

Case 3

When r = 2,

a = bq + r

a = 3q + 2

Cubing on both sides we get,

a³ = (3q + 2)³

Using (a + b)³ = a³ + b³ + 3ab(a + b) we get,

a³ = (3q)³ + (2)³ + 3(3q)(2)[3q + 2]

a³ = 27q³ + 8 + 18q[3q + 2]

a³ = 27q³ + 8 + 54q² + 36q

a³ = 27q³ + 54q² + 36q + 8

a³ = 9(3q³ + 8q² + 4q) + 8

If m = 3q³ + 8q² + 4q

a³ = 9m + 8 → 3

From 1, 2 and 3 we can conclude that the cube of an any positive integer is either of the 9m, 9m+1, 9m+8 for some integer m​.

Hence Proved.