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

We know that any positive integer is of the form 3q or, 3q + 1 or,

3q + 2 for some integer and one and only one of these possibilities can occur.

So, we have following cases:

Case I When n = 3q

In this case, we have

n = 3q, which is divisible by 3

Now,

⇒ n + 2 = 3q + 2,

⇒ n + 2 leaves remainder 2 when divided by 3

⇒ n + 2 is not divisible by 3

Again, n = 3q

⇒ n + 4 = 3q + 4 = 3(q + 1) + 1

⇒ n + 4leaves remainder 1 when divided by 3

⇒ n + 4 is not divisible by 3

Thus, n is divisible by 3 but n + 2 and n + 4 are not divisible by 3.

Case II When n = 3q + 1

In this case, we have

n = 3q + 1

⇒ n leaves remainder 1 when divided by 3

⇒ n is not divisible by 3

Now, n = 3q + 1

⇒ n + 2 = (3q + 1) + 2 = 3(q + 1),

⇒ n + 2 is divisible by 3

Again, n = 3q + 1

⇒ n + 4 = (3q + 1) + 4 = 3q + 5 = 3(q + 1) + 2

⇒ n + 4 leaves remainder 2 when divided by 3

⇒ n + 4 is not divisible by 3

Thus, n + 2 is divisible by 3 but n and n + 4 are not divisible by 3.

Case III When n = 3q + 2

In this case, we have

n = 3q + 2

⇒ n leaves remainder 2 when divided by 3

⇒ n is not divisible by 3

Now, n = 3q + 2

⇒ n + 2 = 3q + 2 + 2 = 3(q + 1) + 1,

⇒ n + 2 leaves remainder 1 when divided by 3

⇒ n + 2 is not divisible by 3

Again, n = 3q + 2

⇒ n + 4 = 3q + 2 + 4 = 3(q + 2)

⇒ n + 4 is divisible by 3

Thus, n + 4 is divisible by 3 but n and n + 2 are not divisible by 3.