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

Any positive integer $n$ 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,$n+2 = 3q+2$ (remainder $2$) and $n+4 = 3q+4 = 3(q+1)+1$ (remainder $1$). Thus,only $n$ 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,$n = 3q+1$ (remainder $1$) and $n+4 = 3q+5 = 3(q+1)+2$ (remainder $2$). Thus,only $n+2$ 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,$n = 3q+2$ (remainder $2$) and $n+2 = 3q+4 = 3(q+1)+1$ (remainder $1$). Thus,only $n+4$ is divisible by $3$.
Conclusion: In all possible cases,exactly one of the numbers $n, n+2, n+4$ is divisible by $3$.