A polynomial of degree 3n has the value 2 at $0, 3, 6, \dots , 3n$, ... A polynomial of degree $3n$ has the value 2 at $0, 3, 6, \dots , 3n$, the value 1 at $1, 4, 7, \dots , 3n-2$ and the value 0 at $2, 5, 8, \dots , 3n-1$. Its value at $3n+1$ is $730$. What is $n$? 
I've tried lots of approaches to this but I can't seem to find the solution.

By using Newton Forward Divided Difference Formula, the polynomial $P_n$ has fractional coefficients and it is equal to:
$$P_n(x)=\sum_{k=0}^{3n}\binom{x}{k}\Delta^kP_n(0).$$
Hence
\begin{align*}
&P_1(x)=2-x+3\binom{x}{3}\\
&P_2(x)=P_1(x)-9\binom{x}{4}+18\binom{x}{5}-27\binom{x}{6}\\
&P_3(x)=P_2(x)+27\binom{x}{7}-81\binom{x}{9}\\
&P_4(x)=P_3(x)+243\binom{x}{10}-486\binom{x}{11}+729\binom{x}{12}.
\end{align*}
Note that $P_4(13)=P_4(3\cdot 4+1)=730$. Hence $n=4$.
