What is the remainder when $1! + 2! + 3! +\cdots+ 1000!$ is divided by $12$?
Solution 1:
If $n\ge 4$, then $4!=24$ divides $n!$ $-$ in particular $12$ divides $n!$ when $\ge 4$.
Thus $$ 1!+2!+\cdots+1000!=1!+2!+3! \!\!\!\!\pmod{12}=9\!\!\!\!\pmod{12}. $$
Solution 2:
Hint: Every term from $12!$ onward is divisible by $12$, so they don't matter.