1. A. Let $n$ be a positive integer, and denote $n!=1 \times 2 \times \cdots \times n$. Then the last digit of $1!+2!+\cdots+10!$ is ( ).
(A) 0
(B) 1
(C) 3
(D) 5
One, 1. A. C.
Notice that, $1!=1, 2!=2, 3!=6, 4!=24$.
It is also known that the last digit of $5!, 6!, 7!, 8!, 9!, 10!$ is 0, and the last digit of $1!+2!+3!+4!$ is 3, so the last digit of $1!+2!+\cdots+10!$ is 3.