4. $1991^{2000}$ divided by $10^{6}$, the remainder is $\qquad$
$$
\begin{aligned}
4.1991^{2000}= & (1+1990)^{2000} \\
= & 1+2000 \times 1990+\frac{1}{2} \times 2000 \times 1999 \times 1990^{2} \\
& +\cdots+1990^{2000} .
\end{aligned}
$$

According to the above expansion, we only need to consider the sum of the first three terms modulo $10^{6}$, so the remainder is 880001.