yanıt: $\boxed{A}$
fermat teoremine göre, $2^{12} \equiv 3^{12} \equiv 4^{12} \equiv 5^{12} \equiv 6^{12} \equiv 1 \pmod{13}$ dür.
Buna göre, $2^{2014}+3^{2014}+4^{2014}+5^{2014}+6^{2014} \equiv 2^{10}+3^{10}+4^{10}+5^{10}+6^{10} \equiv 4^4+(-4)^5+3^5+(-1)^5+(-3)^5 \equiv -1 \equiv 12 \pmod{13}$