Not Lucas' Theorem!

We can first write $n^4$ as: $n^4 = 24\binom{n}{4}+36\binom{n}{3}+14\binom{n}{2}+\binom{n}{1}$ in order to exploit the identity: $\sum_{n=0}^{2014}\binom{n}{k}\binom{2014}{n} = \binom{2014}{k}\cdot 2^{2014-k}.$ Since $819 = 3^2\cdot 7\cdot 13$ , we just have to compute the residue classes $\pmod{7,9,13}$ of: $24\binom{2014}{4}2^{2010}+36\binom{2014}{3}2^{2011}+14\binom{2014}{2}2^{2012}+2014\cdot 2^{2013}.$ Since $\phi(13)=2\phi(9)=2\phi(6)$ , this is the same as calculating the residue classes $\pmod{7,9,13}$ of: $24\binom{2014}{4}2^{6}+36\binom{2014}{3}2^{7}+14\binom{2014}{2}2^{8}+2014\cdot 2^{9},$ or: $(2014\cdot 2013\cdot 2012\cdot 2011)\cdot 2^{6}+6(2014\cdot 2013\cdot 2012)\cdot 2^{7}+7\cdot(2014\cdot 2013)\cdot 2^{8}+2014\cdot 2^{9}.$ The last sum is just $2\pmod{9},5\pmod{7},0\pmod{13}$ , hence the answer is $299$ by the chinese theorem.