Solution: We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2 for some integer q & one and only one of these possibilities can occur
Case I : When n = 3q
In this case, we have,
n=3q, which is divisible by 3
n=3q
= adding 2 on both sides
n + 2 = 3q + 2
n + 2 leaves a remainder 2 when divided by 3
Therefore, n + 2 is not divisible by 3
n = 3q
n + 4 = 3q + 4 = 3(q + 1) + 1
n + 4 leaves a 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 a reaminder 1 when divided by 3
n is not divisible by 3
n = 3q + 1
n + 2 = (3q + 1) + 2 = 3(q + 1)
n + 2 is divisible by 3
n = 3q + 1
n + 4 = 3q + 1 + 4 = 3q + 5 = 3(q + 1) + 2
n + 4 leaves a 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
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 divsible by 3
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
