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Let n be any positive integer and b=3
n =3q+r
where q is the quotient and r is the remainder
0 <r<3
so the remainders may be 0,1 and 2
so n may be in the form of 3q, 3q=1,3q+2
CASE-1
IF N=3q
n+4=3q+4
n+2=3q+2
here n is only divisible by 3
CASE 2
if n = 3q+1
n+4=3q+5
n+2=3q=3
here only n+2 is divisible by 3
CASE 3
If n=3q+2
n+2=3q+4
n+4=3q+2+4
=3q+6
here only n+4 is divisible by 3
Hence it is proved that one and only one among n,n+2,n+4 is divisible by 3 in each case.