A number theory problem by A Former Brilliant Member

$106$ is a multiple of $53$ , so one may think of $1060$ as the solution. but, to be sure, we need to check numbers from $1000$ to $1059$ . If there is one, it would be of form $\overline{10ab}$ , where $0 \leq a \leq 5$ and $0 \leq b \leq 9$ .

then we need the following to be satisfied

$f(\overline{10ab}) \equiv \overline{b10a} \equiv 10^3b+10^2+a \equiv 46(b+1)+1+a \equiv 0 \ (mod \ 53)$

now, we may find $46(b+1)+1 \equiv (mod \ 53)$ for all $0 \leq b \leq 9$ . the result would be as below

$47,40,33,26,19,12,5,51,44,37$

from the list, above, only one has the possibility to be added to a number $0 \leq a \leq 5$ and become equal to $53$ . That would be $51$ , which is achieved when $b=7$ and, then, we add $a=2$ . Therefore, the smallest possible number would be $1027$ .