A problem by Arpit Sah

$12 \mid \overline{ab313ab} \implies 3 \mid \overline{ab313ab} \implies 3 \mid (a + b + 3 + 1 + 3 + a + b) = 7 + 2(a+b)$ So, $\begin{aligned} 7 + 2(a+b) &\equiv 0 \pmod{3}\\ 2(a+b) &\equiv 2 \pmod{3}\\ a+b &\equiv 1 \pmod{3} \end{aligned}$

Also, $12 \mid \overline{ab313ab} \implies 4 \mid \overline{ab313ab} \implies 4 \mid \overline{ab}$ If $a+b=1$ , then there is no way that $4 \mid \overline{ab}$ . If $a+b=4$ , then $a=4,\, b=0 \implies 12 \mid 4031340 = \overline{ab313ab}.$ Hence, $a+b=\boxed{4}$ .