IS. 4 Let $d$ be the remainder of $(196 c)^{2008}$ divided by 97 . Find the value of $d$.
$$
\begin{aligned}
{[196 \times(-3)]^{2008} } & =588^{2008}=(97 \times 6+6)^{2008} \\
& =(97 \times 6)^{2008}+2008 C 1 \cdot(97 \times 6)^{2007} \times 6+\ldots+6^{2008}=97 m+6^{2008}, \text { where } m \text { is an integer. }
\end{aligned}
$$

Note that $2^{5} \times 3=96=97-1 \equiv-1(\bmod 97)$;
$$
\begin{array}{l}
2 \times 3^{5}=486=97 \times 5+1 \equiv 1(\bmod 97) ; \\
\therefore 6^{6}=\left(2^{5} \times 3\right) \times\left(2 \times 3^{5}\right) \equiv-1(\bmod 97) \\
6^{2008}=\left(6^{6}\right)^{334} \times 6^{4} \equiv(-1)^{334} \times 1296 \equiv 97 \times 13+35 \equiv 35(\bmod 97) \\
\therefore d=35
\end{array}
$$