Show that one and only one out of `n ,n+2or ,n+4` is divisible by 3, where `n` is any positive integer.

Explanation

Let a=n, b=n+2 and c=n+4
Order triplet is (a,b,c)=(n,n+2,n+4)
where, n is an integer i.e. n=1,2,3..
`"At n=1 (a,b,c)=(1,1+2,1+4)=(1,3,5)"`
`"At n=2 (a,b,c)=(2,2+2,2+4)=(2,4,6)"`
`"At n=3 (a,b,c)=(3,3+2,3+4)=(3,5,7)"`
`"At n=4 (a,b,c)=(4,4+2,4+4)=(4,6,8)"`
`"At n=5 (a,b,c)=(5,5+2,5+4)=(5,7,9)"`
`"At n=6 (a,b,c)=(6,6+2,6+4)=(6,8,10)"`
`"At n=7 (a,b,c)=(7,7+2,7+4)=(7,9,11)"`
`"At n=8 (a,b,c)=(8,8+2,8+4)=(8,10,12)"`
We observe that each triple consist of ne and only one number which is multiple of 3 i.e. divisible by 3.
Hence one only one out of n,(n+2) and (n+4) is divisible by 3, where n is any positive integer.
Alternate Method
On dividing 'n' by 3 let q the quotient and r be the remainder.
Then n=3q+r, where `0 le r lt 3`
`Rightarrow n=3q+r, where r=0,1,2`
`Rightarrow but n=3q or n=3q+1 or n=3q+2`
CASE I If n=3q, then n is only divisible by 3
but n+2 and n+4 are not divisible by 3.
CASE II If n=3q, then (n+2)=3q+3=3(q+1), which is only divisible by 3.
But n and n+4 are not divisible by 3.
So, in this case, (n+2)is divisible by 3.
CASE III When n=3q+2, then (n+4)=3q+6=3(q+2), which is only divisible by 3.
but n and (n+2) are not divisible by 3.
So,in this case, (n+4) is divisible by 3.
Hence, one and only one out of n,(n+2) and (n+4) is divisible by 3.