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

we can express a number in the form of
a=bq+r where 0 ≤ r <b

Let a be any positive integer and b = 3
a = 3q + r, where 0 ≤ r < 3
possible remainder 0, 1 and 2
If r=0, a = 3q
r=1, a = 3q+1
r=2, a = 3q+2

Case 1:
When a = 3q
Cube on both sides
a^3=(3q)^3
a^3=27q^3
a^3=9(3q^3)
a^3=9m
Where m is an integer such that m = 3q^3

Case 2:
When a = 3q + 1
Cube on both sides
a^ 3 = (3q +1)^3
a^ 3 = 27q^3 + 27q^2 + 9q + 1
a^ 3 = 9(3q^3 + 3q^2 + q) + 1
a^ 3 = 9m + 1
Where m is an integer such that m=(3q^3+3q^2 + q)

Case 3:
When a = 3q + 2
Cube on both sides
a^3 = (3q +2) 3
a^3 = 27q^3 + 54q^2 + 36q + 8
a^3 = 9(3q^3 + 6q^2 + 4q) + 8
a^3 = 9m + 8
Where m is an integer such that m=(3q^3+6q^2 + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

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 .