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, for some integer m.

Answer 1

Let a be any positive integer and it is of the form 3q,3q+1 or 3q+2

Case -1

When a=(3q)³

a³= (3q)³= 27q³

a³=9(3q³)

where m = 3q³

Case-2

when a³ = (3q + 1)³

a³ + 3a²b + 3ab² + b³

⇒(3q)³ + 3(3q)²(1) + 3(3q)(1)² + (1)³

⇒27q³ + 27q² + 9q² + 1

⇒9(3q³ + 3q² ) + 1

9m + 1

where m = 3q³ + 3q

Case - 3

when a³ = (3q + 2)³

a³ = 27q³+54q²=36q+8

⇒ 9(3q³+6q²+4q )+8

⇒9m+8

where m =3q³+6q²+4q

So a³ is of the form 9m,9m+1 or 9m+2

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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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