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