**Solution: **Using Euclid's division lemma any positive integer can be written in the form of a = bq + r where r = 0, 1, 2 ...... and q is quotients.

Let the number which is divisible by 3 be 3q + 0 or 3q + 1 or 3q + 2 where [0 <= r < b]

Now n = 3q or n = 3q + 1 or n = 3q + 2

**Case I,**

When n = 3q .......... (i)

⇒ n = 3(q) where n is divisible by 3

Adding 2 both sides in equ. (i)

We have,

n + 2 = 3q + 2 Where n + 2 is not divisible by 3

Now adding 4 both side in equ. (i)

We have,

n + 4 = 3q + 4 Where n + 4 is not divisible by 3

**Case II **

Taking n = 3q + 1 ........ (ii) where n is not divisible by 3

Adding 2 both sides in equ. (ii)

We have,

n + 2 = 3q + 1 + 2 = 3q + 3

n + 2 = 3(q + 1) where n + 2 is divisible by 3

Now adding 4 both sides in equ. (ii)

n + 4 = 3q + 1 + 4 = 3q + 5 where n + 4 is not divisible by 3

**Case III **

taking n = 3q + 2 .... (iii) where n is not divisible by 3

Adding 2 both sides in equ. (iii)

We have,

n + 2 = 3q + 2 + 2 = 3q + 4 where n + 2 is not divisible by 3

Now adding 4 both sides in equ. (iii)

We have,

n + 4 = 3q + 2 + 4 = 3q +6

n + 4 = 3(q + 2) where n + 4 is divisible by 3

Hence in all these three cases we have seen that either one and only one n or n + 2 or n + 4 is divisible by 3.

*- Answered by ATP Admin* On 17 June 2019:11:28:43 PM