(6) For $n \in \mathbf{N}^{*}$, if $n \cdot 2^{n}+1$ is a multiple of 3, then the set of remainders when $n$ is divided by 6 is $\qquad$.
6 . $\{1,2\}$ Hint: $n \cdot 2^{n}+1=n(3-1)^{n}+1=3 M+(-1)^{n} \cdot n+1$, where $M \in \mathbf{N}$.
Since $3 \mid n \cdot 2^{n}+1$, it follows that $3 \mid (-1)^{n} \cdot n+1$.
When $n=6 k\left(k \in \mathbf{N}^{*}\right)$, $(-1)^{n} \cdot n+1=6 k+1 \equiv 1(\bmod 3)$;
When $n=6 k+1(k \in \mathbf{N})$, $(-1)^{n} \cdot n+1=-6 k \equiv 0(\bmod 3)$;
When $n=6 k+2(k \in \mathbf{N})$, $(-1)^{n} \cdot n+1=6 k+3 \equiv 0(\bmod 3)$;
When $n=6 k+3(k \in \mathbf{N})$, $(-1)^{n} \cdot n+1=-6 k-2 \equiv 1(\bmod 3)$;
When $n=6 k+4(k \in \mathbf{N})$, $(-1)^{n} \cdot n+1=6 k+5 \equiv 2(\bmod 3)$;
When $n=6 k+5(k \in \mathbf{N})$, $(-1)^{n} \cdot n+1=-6 k-4 \equiv 2(\bmod 3)$.
In summary, $3 \mid n \cdot 2^{n}+1$ if and only if $n=6 k+1$ or $n=6 k+2(k \in \mathbf{N})$.