Q:

$a \equiv b \mod n \Rightarrow a$ divided by $n$ gives $b$ as a remainder?

In my lecture notes it is written that $a \equiv b \mod n$ with $a, b, n \in \Bbb Z$ could also be understood in the sense that $a$ divided by $n$ gives $b$ as a remainder. I don't see why this should be true though.
Take $a = 5, \ b = 11$. Obviously $5 \equiv 11 \mod 3$, but $5$ divided by $3$ does not give $11$ as a remainder. What am I missing here?

A:

Well $5\equiv11 \pmod 3$ is also the same as $5\equiv 2 \pmod 3$. 
From the division algorithm, we know
$$p=s\cdot q+r$$
where $r$ is the remainder and is less than $q$. 
Then let $p=5$, $q=3$ and $r=11$ (from your example):
\begin{align}
5&=s\cdot 3+11\\
3s& = -6 \\
s & = -2
\end{align}
Technically, this still works, though some might argue that the condition $q \gt r$ should exist. Think of congruences as a way to find the remainder, but keep in mind that there's flexibility in that definition.

