Easy Problem Easy to get that $$2^n \equiv \begin{cases} 2 & \text{ if } n \text{ odd} \\ 1 & \text{ if } n \text{ even} \end{cases} $$ so it's easy to get the answer $\boxed{n\equiv 1,2 \mod 6}$.

Clearly $n$ cannot be a multiple of $3$, so the possible choices are $n=3k+1, 3k+2$ Case 1: if $n=3k+1$ $3|(3k+1)\cdot(2^{3k+1})+1 \equiv 2^{3k+1}+1 \equiv 0 (mod 3)$ $\Rightarrow 2^{3k+1} \equiv -1 (mod 3)$ $\Rightarrow 2^{k+1} \equiv -1 (mod 3)$ $\Rightarrow (-1)^{k+1} \equiv -1 (mod 3)$ $\Rightarrow k+1$ is odd $\Rightarrow k$ is even therefore $k=2m$ where $m \in \mathbb{Z} \Rightarrow n=3(2m)+1=6m+1$ $2^{3k+1}+1=2^{6m+1}+1$ which is always divisible by $3$. Thus we get $n=6m+1$, $m \in \mathbb{Z}$. The second case is very similar to this case.