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

Answer 1

euclid's lemma states that a=bq +r

let 'a' be any positive integer, 'b' =3 and quotient ='q'.

r=0,1,2

when r=0, a=3q +0 =3q

  a3=[3q]3= 27q3 = 9[3q3] = 9m where m=[3q3]

when r=1, a=3q+1

  a3 = [3q+1]3

  = [3q]3 +[1]3 +3[3q][1]{3q+1}

  = 27q3 +27q2 + 9q + 1

  =9{3q3 + 3q2 + q} + 1  =9m+1 where 'm' =3q3 + 3q2 + q 

when r=2, a= 3q + 2

  a3=[3q+2]3

  =27q3 + 54q2 + 36q + 8

  =9{3q3+ 6q2+4q} +8   =9m +8 hwre 'm' = 3q3+6q2+4q

  hence proved

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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