Is $b\mid a$ standard notation for $b$ divides $a$? Is there a standard way of writing $a$ is divisible by $b$ in mathematical notation?
From what I've search it seems that writing $a \equiv 0 \pmod b$ is one way? But also you can write $b \mid a$ as well (the middle character is a pipe)? And sometimes that pipe is replaced by $3$ vertical dots?
Or is there a way of writing $a$ is a multiple of $b$ which I think means the same thing?
EDIT: thanks for the answers, is there a way to extend this and write something like: $b \mid a$ when $a = k$
 A: I often write that as b divides a
Notation:
$$b \mid a$$
A: *

*Definition: Integer $n$ is a divisible by an integer $d$, when $\exists k \in \mathbb{Z}, n=d\times k$.


*Notation: $d \mid n$


*Synonymous:


*$n$ is a multiple of $d$.


*$d$ is a factor of $n$


*$d$ is a divisor of $n$


*$d$ divides $n$
A: I have seen the following:


*

*$b \mid a$ that is with $\LaTeX$ \mid

*$a = 0 \mod b$ that is with $\LaTeX$ \mod

*$a = 0 \pmod b$ that is with $\LaTeX$ \pmod

*$a \bmod b = 0$ that is with $\LaTeX$ \bmod

*$a \equiv 0\ (b)$

*$a \equiv_{b} 0$


and of course there is


*

*$a = bk$ for some $k \in \mathbb{Z}$


Choose whatever suits you (and your friends or readers) best!
A: Alexander Merkurjev taught  me a long time ago  the ingenious Russian notation  $6 \vdots 2$, which I immediately adopted .
It  pleasantly "rhymes" with  the  equivalent $(6)\subset (2)$
A: $a \equiv 0 \mod b$ and $b \mid a$ are both common, and their use depends on the context.  Given a choice, I use the latter more than the former.
There are others such as $\text{lcm}(a,b)=a$ or $\text{hcf}(a,b)=b$ [or perhaps $\text{gcd}(a,b)=b$ if you prefer] which might also be used when more suitable for the context.  
A: There is also " $a \in b\mathbb Z$ ".
A: COMMENT.- A notation forget in this page is $a\mid\mid b$ which is pertinent many times, very useful really, in number theory. It is used when  a power of a prime, say $p^n$ is the maximal power that divides an integer $M$.
In other words
$$p^n\mid\mid M\iff p^n\mid M \text{  but } p^{n+1}\nmid M$$
