Factorial division and remainders: 100!+102! mod 100 I'm having some issues with factorial division. 
I've been asked to determine the remainder of $11!$ under division by $12$.
My logic was to state that $11! = 1\cdot2\cdot3\cdot4\cdots$ stopping there as $3\cdot4=12$, thus there is a remainder of $0$. Is this correct? Is there a more math-y way of doing this?
Does my logic hold for this example: $100!+102!$ divided by $100$, where $100$ obviously divides $100!$, and $102!=102\cdot101\cdot100\cdots$ thus again there is no remainder?
 A: Yes it is correct, but here exists other less descriptive ways of your  proof.
$$\left(\forall n \in \mathbb{N}^{+}\right)
\left(\forall a \mid n\right)
\left(\exists k \in \mathbb{Z}\right)
\left( n = ak \right)$$
So, $\left(100 \mid 102! \wedge 100\mid 100!\right) \Rightarrow \left(\exists k',k'' \in\mathbb{Z}\right)\left(100k' = 100! \wedge 100k'' = 102!\right)$, in accordance with this
$$100! + 102! = 100 k' + 100 k'' = 100(k'+k'') \equiv 0 \pmod{100}$$
You can also provide it other way.
$$\begin{split}
\left(100 \mid 100! \wedge 100 \mid 102!\right) &\Longrightarrow \left(  100! \equiv 0 \mod{100} \wedge 102! \equiv 0 \mod{100}\right)\\ &\Longrightarrow  100! + 102! \equiv 0 \pmod{100}
\end{split}$$
But it's just other words for that.

We should note, that $100 \mid 100! \wedge 100 \mid 102!$. In fact you didn't have to write so much to prove it ($102! = 102 \cdot 101 \cdot 100 \cdot 99! \Rightarrow 100 \mid 102!$)
what if I ask about 1234!? We should note simple fact here.
$$\left(\forall n \in \mathbb{N}^{+}\right)\left(\forall a' \in \mathbb{N}^{+}\right) \left(a' \leq n \Longrightarrow a' \mid n!\right)$$
I think it's obvious, because $n! = 1 \cdot 2 \cdot 3 \cdot ... \cdot n$. Problematic can be question, if $291 \mid 100!$? But we can note $291 = 97 \cdot 3$, and $3,97 < 100 \wedge 3 \neq 97 $ so yes.
