TITLE: Modulo of a Sum/Summation
QUESTION [0 upvotes]: Let me define $a~mod~N \triangleq a/N$. It is well known that, 
$(a+b)/N = (a/N + b/N)/N$.
Can some one clarify how to translate this to simplify, $(a + b + c)/N$?
From (Link 1) and (Link 2), it is said that, 
$(a + b + c)/N = (a/N + b/N + c/N)/N$ but this is not clear to me. From the case of $a + b$, of I set $b = b +c$, I should get, 
$$\begin{align*}
  \left( {a + b + c} \right)/N & = \left( {a/N + (b + c)/N} \right)/N \hfill \\
   & = \left( {a/N + \left( {b/N + c/N} \right)/N} \right)/N. \hfill \\ 
\end{align*}$$ 
This does not explain the linked results. 
Can this also be used for functions $k_p(x)$,  
$$\left( {\sum\limits_{p = 1}^P {{k_p}\left( x \right)} } \right)/N.$$
Your help is appreciated.

REPLY [0 votes]: Your result is actually true as well.
Their result uses that the "modulo" operation squares to itself,
i.e. (a/N) = (a/N)/N for all a.
You should think of modulo in a different way.
You have your normal ring of numbers (say $\mathbb{Z}$ or $\mathbb{R}$) and then you have the modulo operation $\varepsilon_{n}: \mathbb{Z} \to \mathbb{Z}_{n}= \{[0]_{n}, [1]_{n}, [2]_{n}, ..., [n-1]_{n}\}$.
$\mathbb{Z}_{n}$ is a new ring with only n elements. Each element is an equivalence class $[a]_{n}$ of elements $a \in \mathbb{Z}$ which have the same reminder dividing by n. So $[a]_{n}$ and $[a+n]_{n}$ are actually the
same element in the ring $\mathbb{Z}_{n} = \{[0]_{n}, [1]_{n}, [2]_{n}, ..., [n-1]_{n}\}$.
What your "/N" is is essentially applying $\varepsilon$ to get the equivalence class modulo n and then applying the inclusion $i: \mathbb{Z}_{n} \to \mathbb{Z}$ by choosing a representative, i.e. the smallest non-negative representative. (Note that this inclusion depends of the chose of a repr.)
Upshot: You can take as many modulo as you want as long as you take one at the end. While being in the modulo ring adding and subtracting n to the repr. a doesn't change the object $[a]_{n}$
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