Question

Using Euclid's division lemma, show that the cube of any positive integers is of the form 9m,9m+1,9m+8 for some integer m

Answer 1

Answer:

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Step-by-step explanation:

Let 'a' be any positive integer and b=9

then by using eucild's  division lemma

a=9q+r      where,

q≥0 & 0≤r≤9

∴a=9q (or) 9q+1 (or) 9q+2 (or) 9q+3 (or) 9q+4 (or) 9q+5 (or) 9q+6 (or) 9q+7 (or) 9q=8

let 'a'=3n

C.O.B.S

a³=(3n)³

a³=27n³

a³=9(3n³)                      [∵3n³=m]  

a³= 9m

a=3n+1

C.O.B.S

a³=(3n+1)³

a³=27n³+27n²+9n+1

a³=9(3n³+3n²+n)+1

a³=9m+1

a=3n+2

C.O.B.S

a³=(3n+2)³

a³=27n³+54n²+36n+8

a³=9(3n³+6n²+4n)+8

∴Cube of any number can be written in the form of 9m(or)9m+1(or)9m+8

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