Prove that one and only one out of $n, (n + 2)$ and $(n + 4)$ is divisible by 3, where $n$ is any positive integer.

To do: 

We have to prove that one and only one out of $n, (n + 2)$ and $(n + 4)$ is divisible by 3, where $n$ is any positive integer.

Solution:

Let $a = n, b = n + 2$ and $c =n + 4$

Order triplet is $(a, b, c) = (n, n + 2, n + 4)$....…(i)

where, $n$ is any positive integer

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 triplet consists of one and only one number which is a multiple of 3.

Hence, one and only one out of $n, (n + 2)$ and $(n + 4)$ is divisible by 3, where, $n$ is any positive integer.