$\begin{aligned}
& 7^{103}=7\left(7^{102}\right)=7(343)^{34}=7(345-2)^{34} \
& 7^{103}=23 \mathrm{K}_1+7.2^{34}
\end{aligned}Now7.2^{34}=7.2^2 \cdot 2^{32}\begin{aligned}
& =28 \cdot(256)^4 \
& =28(253+3)^4 \
& \therefore 28 \times 81 \Rightarrow(23+5)(69+12) \
& 23 \mathrm{K}_2+60 \
& \therefore \text { Remainder }=14
\end{aligned}$