8. Let $A=100101102103 \cdots 798799$ be a 2100-digit positive integer, formed by sequentially connecting the three-digit numbers from 100 to 799. Then the remainder when $A$ is divided by 126 is $\qquad$
8.91.

Notice that, $126=2 \times 9 \times 7$,
$$
\begin{array}{l}
A \equiv 1(\bmod 2) \text {, } \\
A \equiv 100+101+\cdots+799 \equiv 899 \times 350 \\
\equiv 1(\bmod 9) \text {, } \\
A=100 \times 1000^{699}+101 \times 1000^{698}+\cdots+ \\
798 \times 1000+799 \\
\equiv 100 \times(-1)^{699}+101 \times(-1)^{698}+\cdots+ \\
798 \times(-1)+799 \\
\equiv-100+101-102+103-\cdots-798+799 \\
\equiv 350 \equiv 0(\bmod 7) \text {. } \\
\end{array}
$$

Therefore, $A \equiv 91(\bmod 126)$.