Question

use euclids division Lemma , to show that the cube of any positive integer is of the form 9 M, 9 M + 1 or 9 M + 8 for some integer m​

Answer 1

Explanation:

let a= cube of any positive integer

let b=3

Applying Euclid division lemma ,

a=3q+r

now,since b=3 then r=0,1,2

so, we have possible outcomes of,

CASE-1- a=3q+0

CASE-2- a=3q+1

CASE-3- a=3q+2

Now, In CASE-1-

a=3q

a^3=27q^3

a^3= 9×(3q^3)

a^3=9m------------- where m=3q^3

CASE-2-

a=3q+1

a^3=27q^3 + 27q^2 + 9q+1

a^3=9 (3q^3+3q^2+q)+1

a^3=9m+1----------------where m=(3q^3+3q^2+q)

CASE-3-

a=3q+2

a^3=27q^3+ 54q^2 +18q + 8

a^3=9 (3q^3+6q^2+2q)+8

a^3=9m+8----------------where m=(3q^3+6q^2+2q)

Hence,we can show any cube of positive integer in form of 9m,or 9m+1,or9m+8

Answer 2

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 .