Question

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

Answer 1

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

 

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 =  (3q)3 = 27q3

9(3q3) = 9m

Case 2: When a = 3q + 1,

a 3 = (3q +1) 3 

a 3 = 27q 3 + 27q 2 + 9q + 1 

a 3 = 9(3q 3 + 3q 2 + q) + 1

a 3 = 9m + 1 

Where m is an integer such that m = (3q 3 + 3q 2+ q) 

Case 3: When a = 3q + 2,

a 3 = (3q +2) 3 

a 3 = 27q 3 + 54q 2 + 36q + 8 

a 3 = 9(3q 3 + 6q 2 + 4q) + 8

a 3 = 9m + 8

Where m is an integer such that m = (3q 3 + 6q 2+ 4q) 

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Answer 2

Step by step Explanation :-

• Let the positive integer be a which when divided by 3 gives q as quotient and r as remainder.

by Euclid's division lemma

a=bq+r

a=3q+r

where r=0,1,2

then,

a=3q

or

a=3q+1

or

a=3q+2

now,

• CASE i

a=3q

a³=(3q)³

a³=27q³

a³=9m( where m=3q³)

• CASEii........ using identity

(a+b)³=a³+3a²b+3ab²+b³

a=3q+1

a³=(3q+1)³

a³=27q³+9q²+9q+1

a³=9m(3q³+q²+q)+1

a³=9m+1

• CASE iii

a=3q+2

a³=(3q+2)³

a³=27q³+27q²+36q+8

a³=9m(where m=3q³+3q²+4q)+8

a³=9m+8

Hence , the cube of any +ve no. is of the form 9m , 9m +1 or 9m+8

is shown !!