Show that one and only one out of n,( n+2) and ( n +4) is divisble by 3, where n is any positive interger.

Explanation

When , n is dividied by 3 , let q and r be the quotient and remainder respectively .
`therefore " "` n = 3q + r
where , `0 le r lt 3 ` i.e., r = 0 or r =1 or r =2
(i) When , r = 0 then
n = 3q ,
which is divisible by 3 .
n + 1 = 3q +1
which leaves a remainder 1 when divided by 3.
i.e., (n +1) is not divisible by 3 .
n +2 = 3q +2
which leaves a remainder 2 when divided by 3 .
i.e., (n +2) is not divisible by 3 .
So , only n = 3q is divisible by 3 when r = 0 . `" " .... (1)`
(ii) When r = 1 , then
n = 3q +1
which is not divisible by 3 . `" " (because` it leaves a remainder 1)
n + 1 = 3q +2
which is not divisible by 3. `" " (because` remainder= 2)
n + 2 = 3q + 3 = 3(q+1)
which is divisible by 3 .
So , only n +2 is divisible by 3 when r= 1 . `" "....(2)`
(iii) When , r=2 then
n = 3q +2
which is not divisible by 3 . `" " (because ` remainder =2)
`implies " " n + 1 = 3q + 3 = 3(q +1)`
which is divisible by 3.
n +2 = 3q + 4 = 3q + 3 + 1 = 3(q+1) + 1
which is not divisible by 3 `" " (because` remainder =1)
So , only n+1 is divisible by 3 when r =2 . `" " .... (3)`
Hence , from equation (1) , (2) and (3) , we can say that one and only one out of n , (n +1) and (n +2) is divisible by 3.