Math, asked by Skysweetwini6278, 1 year ago

How to show that the cube of any positive integer is in the form of 9m, 9m+1 or 9m+8

Answers

Answered by shadowsabers03
0

     

Write down first a few perfect cubes. First 15 perfect cubes are given below:

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375...

Divide each terms by 9 and write down the remainders thus obtained:

1, 8, 0, 1, 8, 0, 1, 8, 0, 1, 8, 0, 1, 8, 0...

 

The remainders are recurring by the numbers 0, 1 and 8.

So the cube of any positive integer is in the form 9m, 9m + 1 and 9m + 8!!!

Hence proved!!!

Plz ask me if you have any doubt on my answer.

Thank you...

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faseeh15: bad answer
Answered by Anonymous
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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