I4.4 Let $D$ be the sum of the last three digits of $2401 \times 7^{C}$ (in the denary system). Find the value of $D$.
$\begin{array}{l}2401 \times 7^{C}=7^{4} \times 7^{12}=7^{16}=\left(7^{2}\right)^{8}=49^{8}=(50-1)^{8} \\ =50^{8}-8 \times 50^{7}+\ldots-56 \times 50^{3}+28 \times 50^{2}-8 \times 50+1 \\ \equiv 28 \times 2500-400+1(\bmod 1000) \\ \equiv-399 \equiv 601(\bmod 1000) \\ D=6+0+1=7 \\ \text { Method } 22401 \times 7^{C}=7^{4} \times 7^{12}=7^{16} \\ 7^{4}=2401 \\ 7^{8}=(2400+1)^{2}=5760000+4800+1 \equiv 4801(\bmod 1000) \\ 7^{16} \equiv(4800+1)^{2} \equiv 48^{2} \times 10000+9600+1 \equiv 9601(\bmod 1000) \\ D=6+0+1=7 \\\end{array}$