Convoluted Congruence

Let us try to interpret what each congruence means.

$1 \equiv 2017 \: (\!\! \mod x)$

This means that when $2017$ is divided by an integer $x$ , a remainder of $1$ is left. Or, in math-ese, $2017 = kx + 1$ for some integer $k$ . We can deduce from here that $kx = 2016$ , so that $x$ must be a factor of $2016$ .

Similarly, we can derive that same piece of information from the two remaining systems ( since $2018 = jx + 2$ for some integer $j$ , and $2019 = mx + 3$ for some integer $m$ ). This allows us to investigate on the factors of $2016$ .

$2016 = 2^5 \cdot 3^2 \cdot 7$ , which means there are $36$ factors, including $2016$ itself and $1$ . Now, we will remove the values of $x$ which may be factors of any of $2017$ , $2018$ , or $2019$ as well. Since $2017$ is prime, there's no worry that any $x$ from the factors of $2016$ will divide it evenly.

As for $2018$ , we know that $\gcd (2016, 2018) = 2$ , so we remove $2$ from the list.

As for $2019$ , we know that $\gcd (2016, 2019) = 3$ , letting us remove $3$ now.

Having removed $1$ , $2$ , $3$ , and $2016$ from the $36$ factors leaves us with $\boxed{32}$ .