What does this math symbol mean $|$? I came across this symbol in my discrete mathematics course. 
For example: $6 | n(n+1)(n+2)$ 
 A: $a|b$ means $a$ divides $b$, that is we can find $k \in \mathbb{Z}$ such that $b=ak$.
$$6|n(n+1)(n+2)$$ means we can find $k \in \mathbb{Z}$ such that $n(n+1)(n+2)=6k.$
A remark about the formula is that among $3$ consecutive numbers there is $a$ multiple of $3$ and between $2$ consecutive numbers, there is an even number.
A: The notation $a | b$ refers to $a$ divides $b$, i.e. there exists $w \in \mathbb{Z}$ such that
$$
a \cdot q=b
$$
A: Jonathan's answer and ajotaxe's comments are correct.  $a\mid b$ means "$a$ divides $b$" (or "$a$ divides into $b$", or sometimes "$a$ divides evenly into $b$") which means there is an integer $c$ so that $b = c*a$ or in other words so $\frac ba$ is an integer.
So in you example the statement "$6|n(n+1)(n+2)$ for all $n \in \mathbb N$" is the statement that "For any natural number $n$, we will have that $6$ will divide into $n(n+1)(n+2)$".  For example: if $n = 13$ then $13*14*15 =  2730$ and $6$ divides into $2730$ evenly because $2730 = 6*455$. Or in other words $\frac {13*14*15}{6} = 455\in \mathbb Z$
There is a small caveat.  For $a|b$ it isn't necessary that $a$ and $b$ themselves have to be integers;  just that  $\frac ba$ is an integer.  So for example.   $\frac 18\mid \frac 34$ becasue $\frac 34 = 6*\frac 18$ and $6$ is an integer; of $\frac 23 \mid -8$ because $-8 = -12*\frac 23$ and $-12$ is an integer.  But $\frac 13 \not \mid \frac 34$ because $\frac {\frac 34}{\frac 13} = \frac 94;$ or $\frac 34 = \frac 94*\frac 13$ and $ \frac 94$ is not an integer.
