I4.4 Find the value of $1+3^{c}+\left(3^{c}\right)^{2}+\left(3^{c}\right)^{3}+\left(3^{c}\right)^{4}$ divided by $1+3+3^{2}+3^{3}+3^{4}$, denoted as $d$.
Determine the remainder $d$ of $1+3^{c}+\left(3^{c}\right)^{2}+\left(3^{c}\right)^{3}+\left(3^{c}\right)^{4} \quad$ divided by $1+3+3^{2}+3^{3}+3^{4}$.
$$
\begin{array}{l}
1+3+3^{2}+3^{3}+3^{4}=1+3+9+27+81=121 \\
1+3^{c}+\left(3^{c}\right)^{2}+\left(3^{c}\right)^{3}+\left(3^{c}\right)^{4}=1+9+81+729+6561=7381=121 \times 61
\end{array}
$$

The remainder $d=0$