Question

use euclids division lemma to show that the cube of any positive integer is of the form 9m,9m plus 1or 9m plus 8??

Answer 1

Thus so

LET ANY POSITIVE INTEGER BE n

THEN BY EUCLID S DIVISION LEMMA

➡ D = BQ+ R

. D. =.( 3 Q + 0 ) ³
. 27 Q ³
. 9 ( 3 Q ³ )
. 9 M

. D = ( 3 Q + 1 ) ³
. 27 Q ³ + 1 + 9Q ( 3Q + 1 )
. 27 Q ³ + 1 + 27 Q ² + 9 Q
. 9 ( 3 Q ³ + 3 Q ² + Q ) + 1
. 9 M + 1

. D = ( 3 Q + 2 ) ³
. 27 Q ³ + 8 + 18 Q ( 3 Q + 2 )
. 27 Q ³ + 8 + 54 Q ² + 36 Q
. 9 ( 3 Q ³ + 6 Q ² + 4 Q ) + 8

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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.