find the least positive residue of $1!+2!+3!+...+100!$ modulo each of the following integers I am trying to find the least positive residue of $1!+2!+3!+...+100!$ modulo each of the following integers:
a) $2$
b) $7$
c) $12$
d) $25$ 
and I am stuck on how to do this. I know that you have to set $n \geq 1$ for a) and state $n!$ is divisible by $1 = 1$ but I do not know where these numbers come from. 

$$1!+2!+\dots 100!\\ \equiv 1! +2!+3\times 2!+ 4\times 3\times  2!...\\ \equiv 1!+0+3\times 0+ 4\times 3\times  0...\equiv 1!\pmod 2$$
Because $2!$ is congruent to $0$ $\pmod 2$. As André Nicolas pointed out in the comments, $4!\equiv 0\pmod {12}$ which means that $5!$ which is $5\times 4!$ is congruent to $0$ as well. See if you can generalize this strategy to other moduli as well.
EDIT: Lets do the $\pmod {12}$ example. 
$$1!+2!+\dots 100!\\ \equiv 1! +\ldots+4!+5\times 4! + 6\times 5\times 4!+7\times 6\times 5\times 4!+...\\ \equiv 1! +\ldots+24+5\times 24 + 6\times 5\times 24+7\times 6\times 5\times 24+... \pmod {12}$$ Because $4!=24$. $24=12\times 2$ which is congruent to $0\times 2=0$ $\pmod{12}$. So... $$1! +\ldots+24+5\times 24 + 6\times 5\times 24+7\times 6\times 5\times 24+..\\ \equiv 1!+2!+3!+0+5\times 0+6\times 5\times 0\ldots\\ \equiv 1!+2!+3! \pmod{12}$$ and $1!=1,2!=2,3!=6$ so $1!+2!+3!=1+2+6=9$
