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

(N/A) Let $n$ be any positive integer. By Euclid's division lemma,$n$ can be expressed in the form $3q, 3q+1,$ or $3q+2$ for some integer $q \ge 0$.
Case $1$: If $n = 3q$,then $n$ is divisible by $3$. In this case,$n+2 = 3q+2$ and $n+4 = 3q+4 = 3(q+1)+1$,neither of which is divisible by $3$.
Case $2$: If $n = 3q+1$,then $n+2 = 3q+1+2 = 3q+3 = 3(q+1)$,which is divisible by $3$. In this case,$n = 3q+1$ and $n+4 = 3q+1+4 = 3q+5 = 3(q+1)+2$,neither of which is divisible by $3$.
Case $3$: If $n = 3q+2$,then $n+4 = 3q+2+4 = 3q+6 = 3(q+2)$,which is divisible by $3$. In this case,$n = 3q+2$ and $n+2 = 3q+2+2 = 3q+4 = 3(q+1)+1$,neither of which is divisible by $3$.
Thus,in all possible cases,exactly one of $n, n+2,$ or $n+4$ is divisible by $3$.