1. Let $a_{n}$ denote the last two digits of $7^{n}$, then the last three digits of $a_{1}+a_{2}+\cdots+a_{2005}$ are ( ).
(A) 109
(B) 107
(C) 103
(D) 101
-、1.B.
Since $n$ is a positive integer, the last two digits of $7^{n}$ cycle through $07, 49, 43, 01$. Also, $2005=4 \times 501+1$, so
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2005} \\
=501 \times(7+49+43+1)+7=50107 .
\end{array}
$$