We applied Euclid Division algorithm on n and 3.

a = bq + r on putting a = n and b = 3

n = 3q + r, 0 < r < 3

So,

n = 3q (I)

n = 3q + 1 (II)

n = 3q + 2 (III)

Case - I: When n = 3q

In this case, we have

n = 3q, which is divisible by 3

Now, n = 3q

n + 2 = 3q + 2

n + 2 leaves remainder 2 when divided by 3

Again, n = 3q

n + 4 = 3q + 4 = 3(q + 1) + 1

n + 4 leaves 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 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.