$2abc = cab$

take the number $X=\overline{a_n \dots a_1}$ , then

$2\times \overline{a_n \dots a_1}= \overline{a_1 a_n\dots a_2} \implies 19(\sum_{j=0}^{j=n-2}10^{j}a_j)=(10^n-2)a_1$

so, we need to find the lowest $n$ , for which $19 | 10^n-2$ or $10^n\equiv 2 \ (mod \ 19)$ .

$10^n\equiv 2 \ (mod \ 19) \implies 5.10^{n-1}\equiv 1\ (mod \ 19) \implies 10^{n-1}\equiv 5^{-1} \equiv 4\ (mod 19)$

$10^{n-1} \equiv 4 \ (mod \ 19) \implies 25.10^{n-3} \equiv 1 \ (mod \ 19) \implies 10^{n-3} \equiv 25^{-1} \equiv 16 \ (mod \ 19)$

$10^{n-3} \equiv 16 \implies 5^4.10^{n-7} \equiv 1 \implies 10^{n-7} \equiv 9 \ (mod \ 19)$

$10^{n-7} \equiv 9 \equiv -10 \implies 10^{n-8} \equiv -1 \ (mod 19)$ .

So, we need to find the smallest number of form $10^k+1$ that is divisible by $19$ . Luckily, for $k=9$ , $10^k+1$ is divisible by $19$ and we don't have to go through more trouble. So, $n-8=9 \implies n=17$ . We now go back to

$19(\sum_{j=0}^{j=n-2}10^{j}a_j)=(10^n-2)a_1 \implies \sum_{j=0}^{j=n-2}10^{j}a_j=\frac{(10^{17}-2)}{19}a_1 \implies \frac{X-a_1}{10}=\frac{(10^{17}-2)}{19}a_1$

since $a_1=2.a_{16}$ , $a_1$ needs to be even and since its multiplication by $\frac{(10^{17}-2)}{19}$ determines the number, without its last digit $a_1$ , we take the minimum $a_1=2$ . then, we only need to know the last two digits of $\frac{(10^{17}-2)}{19}$ to know the last three digits of $X$ . for that, we calculate

$(10^{17}-2).(19)^{-1} \equiv (-2).79 \equiv 42 \ (mod \ 100) \implies a_3=2.4,a_2=2.2$

The smallest n-parasitic numbers are also known as Dyson numbers, after a puzzle concerning these numbers posed by Freeman Dyson. In the present case, n=2 and A=105263157894736842, the derivation of which may be read here: https://en.wikipedia.org/wiki/Parasitic_number