Show that the cube of any positive integer of the form 6q +r,q is an integer and r is equal to zero, 1, 2, 3, 4, 5 is also a form of 6m+r

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Solution

let a be any +ve integer
b=6

by euclid 's division lemma,

a=bq+r, 0=< r < b
a= 6q+r,0=< r

when

r=0,a = 6q= (6q)3 ---> a3 = 216q3 =6(36q3)=6q(where m is = 6q3 )

by similar manner you can prove for

r=1,2,3,4,5
and you will get the proof

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Show that the cube of any positive integer of the form 6q +r,q is an integer and r is equal to zero, 1, 2, 3, 4, 5 is also a form of 6m+r

let a be any +ve integer b=6 by euclid 's division lemma, a=bq+r, 0=< r < b a= 6q+r,0=< r when r=0,a = 6q= (6q)3 ---> a3 = 216q3 =6(36q3)=6q(where m is = 6q3 ) by similar manner you can prove for r=1,2,3,4,5 and you will get the proof

let a be any +ve integer
b=6

by euclid 's division lemma,

a=bq+r, 0=< r < b
a= 6q+r,0=< r

when

r=0,a = 6q= (6q)3 ---> a3 = 216q3 =6(36q3)=6q(where m is = 6q3 )

by similar manner you can prove for

r=1,2,3,4,5
and you will get the proof