How to find the remainder when the following series is divided by 12?

Question

$1! + 2! + 3!+\cdots + 99! + 100!$

I am not getting any idea on how to solve this problem. I know that modular arithmetic should be used but not getting how to start off with the solution. Please give me some hint on how to approach this question. Thanks in advance.

Answer 1

Hint: Note that $4!\equiv 0\pmod{12}$.

Answer 2

Hint: Most elements in the sum are divisable by $12$

Answer 3

Hint $\ n(n\!-\!1)\mid n!\mid (n\!+\!1)!\mid (n\!+\!2)!\mid \cdots\ $ for $\, n>1$

Answer 4

Since $4!$ is congruent to $0$ (mod $12$) then any multiple of $4!$ is congruent to $0$ (mod $12$). So we need only look at the first 3 terms, and since each of the first 3 terms is congruent to themselves (mod $12$), then the addition of all the terms in (mod $12$) is:

$(1! + 2! + 3! + 0 + 0 +...+ 0)$(mod 12). So the remainder should be 9.