4. What is the sum of the last two digits of the integer $1!+2!+3!+\cdots+2005$ !?
(A) 3
(B) 4
(C) 5
(D)
(E) 7
4. Ans: B
One easily checks that for $n \geq 10, n!$ will end with two zeros. Also,
$$
\begin{aligned}
& 1!+2!+3!+4!+5!+6!+7!+8!+9! \\
= & 1+2+6+24+120+720+5040+40320+362880 \\
= & 409113
\end{aligned}
$$

Thus the sum of the last two digits is $1+3=4$.