Divisibility Rules of 2019 What is the divisibility rule or way to find that a number is divisible by 2019 and we can't divide the number by 2019 to test it? and how do we prove that the rule works for 2019?
Help is appreciated!  
 A: $2019 = 3(673)\,$ so it suffices by CRT to compute the remainders mod $3\,$ & $\,673$.
$\!\!\bmod 3\!:\,$ the remainder is congruent to the digit sum (as in casting out nines).
$\!\!\bmod 673\!:\  10^{\large 14}\equiv 8\,$ so we can use that to work in chunks of $\,14\,$ decimal digits, e.g.
$\ n = 8100000000000025= 81(10^{\large 14})+25 \equiv 81(8)+25\equiv 673\equiv \color{#0a0}0$
$ $ By $ $ Easy $ $ CRT: $\,\ \ \ \begin{align} &n\equiv \color{#0a0}a\pmod{\!673}\\ &n\equiv\color{#c00} b\pmod{\!3}\end{align}\iff\, n\equiv \color{#0a0}a + 673(\color{#c00}b\!-\!\color{#0a0}a)\,\pmod{\!2019}$
e.g. above $\,n\equiv 8\!+\!1+\!2\!+\!5\equiv\color{#c00} 1\pmod{\!3}\,$ so $\ n\equiv \underbrace{\color{#0a0}0+673(\color{#c00}1\!-\!\color{#0a0}0)}_{\large 673}\,\pmod{\!2019}$
For some numbers this may be faster than the universal divsiibility test, which is essentially a modular form of the long division algorithm that ignores the quotients. 
A: In the spirit of the usual rule for $7$, which checks whether a number is divisible but does not give the remainder if not, given a number $n$ to check, you can delete the last digit of $n$ and add $202$ times that digit to the result.  You can keep going until you get down to four digits and there are not many multiples of $2019$ with four digits.  
It works because if the last digit is $d$ we are saying that $n$ is divisible by $2019$ if and only if $n+2109d$ is, but $n+2019d$ ends in $10$ so we can divide by $10$.  
As an example, if $n=954987$ we next get 
$$95498+202\cdot 7=96912\\
9691+202\cdot 2=10095\\
1009+5\cdot 202=2019$$
so $954987$ is divisible by $2019.$  In fact it is $2019 \cdot 473$
