Question

Using Euclid's division lemma show that cube of any positive integer is of form 9m ,9m+1 or 9m+8. With proper explanation of each step.

Answer 1

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Answer 2

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Solution
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Let n be an arbitrary position integer.
On dividing n by 3 , let q be the quotient and r be the remainder .

so , by Euclid's division lemme, we have

n = 3q + r , where 0 ≤ r < 3.
=> n^3 = (3q + r )^3 = 27q^3 + r^3 + 9qr (3q + r)

=> (27q^3 + 27q^2r + 9qr^3) + r^3 ......( ¡ )
where 0 ≤ r < 3.


CASE I – when r = 0.
putting r = 0 in ( ¡ ), we get
n^3 = 27q^3=9(3q^3)=9m ,
where m = 3q^3 is an integer.


CASE II – when r = 1
putting r = 1 in ( ¡ ) , we get
n^3 =(27q^3+27q^2+9q)+1=9q(3q^2+3q+1)+1

= 9m + 1 , where m = q(3q^2 + 3q + 1) is an integer.

CASE III – when r = 2.
putting r = 2 in ( ¡ ), we get
n^3 = (27q^3+54q^2+36q)+8 =9q(3q^2+6+4)+8

9m + 8 , where m = q(3q^2 + 6q + 4) is an integer.

Hence , the cube of any positive integer is of the form 9m or (9m + 1 ) or ( 9m + 8 ) for some integer m.

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