Question

use euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8.

Answer 1

If x Ξ 0 mod 3, then there is some positive integer k such that x = 3k. Therefore 
x³ = (3k)³ = 27k³ = 9(3k³) which is of the form 9m with m=3k³.

If x Ξ 1 mod 3, then there is some nonnegative integer k (could be 0) such that x = 3k+1. Then 
x³ = (3k+1)³ = (3k+1)²(3k+1) = (9k²+6k+1)(3k+1) = 27k³+27k²+9k + 1 = 9(3k³+3k²+k) + 1 which is of the form 9m+1 with m=3k³+3k²+k

If x Ξ 2 mod 3, then there is some nonnegative integer k (could be 0) such that x = 3k+2. Then 
x³ = (3k+2)³ = (3k+2)²(3k+2) = (9k²+12k+4)(3k+2) = 27k³+54k²+36k + 8 = 9(3k³+6k²+4k) + 8 which is of the form 9m+8 with m=3k³+6k²+4k.

Answer 2

Step-by-step explanation:

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ r = 0,1,2 .

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

Where m is an integer such that m =

Case 2: When a = 3q + 1,

a = (3q +1) ³

a = 27q ³+ 27q ² + 9q + 1

a = 9(3q ³ + 3q ² + q) + 1

a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .

Case 3: When a = 3q + 2,

a = (3q +2) ³

a = 27q³ + 54q² + 36q + 8

a = 9(3q³ + 6q² + 4q) + 8

a = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Hence, it is proved .