Distribution of $\bmod$ on the $+$ operator We know that $(a*b) \pmod n \equiv  (a \pmod n * b \pmod n) \pmod n$.
What other distributive attributes of the mod exist? Specifically when we have:
$(a + b + c) \pmod n \equiv ?$
 A: Of course
$$(a+b+c)\equiv(a\pmod n+b\pmod n+c\pmod n)\pmod n$$
update:
Actually, we can prove the general case, which is
$$\left(\sum_{i=1}^na_i\right)\equiv\sum_{i=1}^n\left(a_i\pmod n\right)\pmod n$$
And this can be proved using induction.
For the base case, we need to prove
$$(a_1+a_2)\equiv(a_1\pmod n+a_2\pmod n)\pmod n$$
Move the right side to left then we need to prove
$$(a_1+a_2-a_1\pmod n-a_2\pmod n)\equiv0\pmod n$$
Assume $a_1=d_1\cdot n+r_1$, $a_2=d_2\cdot n+r_2$, where $0\leq r_1, r_2<n$ are the residue modulo $n$. It is obvious that $r_1=a_1\pmod n$ and $r_2=a_2\pmod n$.
As a result, $a_1-a_1\pmod n=d_1\cdot n$ and $a_2-a_2\pmod n=d_2\cdot n$. Therefore, it holds that
$$(a_1+a_2-a_1\pmod n-a_2\pmod n)=(d_1\cdot n+d_2\cdot n)\equiv0\pmod n$$
So the base case is proved. And the induction step is trivial,
\begin{eqnarray}
\left(\sum_{i=1}^na_i\right)&=&\left(\sum_{i=1}^{n-1}a_i\right)+a_n\\
&\equiv&\left(\left(\sum_{i=1}^{n-1}a_i\right)\pmod n+a_n\pmod n\right)\pmod n\;\;(\textrm{by base case})\\
&\equiv&\left(\sum_{i=1}^{n-1}\left(a_i\pmod n\right)+a_n\pmod n\right)\pmod n\;\;(\textrm{by induction hypothesis})\\
&\equiv&\sum_{i=1}^n\left(a_i\pmod n\right)\pmod n
\end{eqnarray}
