Question

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

Answer 1

Step-by-step explanation:

Let a and b are two positive integer. Then by euclids division lemma. a =bq +r where b=3

So we can find r = 0, 1, 2

:(3 cubeq) = 27q = 9(3q) m=3q=9m

:3q+1 =9(3q)+1, m =3q,=9m+1

:(3q+2cube)=9(q)+8 m=q=9m+8

So our positive integer are 9m, 9m+1,9m+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 .

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THANKS.