Open Contests

Let the number be $aba$ and let the number we get by adding $2016$ be $cddc$. Clearly $c$ can only be $2$ or $3$.

If $c=2$ then by the ones digit the only possibility is $a=6$, and we have a carry from the ones to the tens digit. By the tens digit then $b+1+1=d$ or $b+1+1=d+10$. The second option is impossible, since by the hundreds digit we can only have $d=6$, if there is no carry from the tens to the hundreds digit, and $d=7$, if there is a carry from the tens to the hundreds digit. Thus $b+2=d$. Then there is no carry to the hundreds digit, hence $d=6$ and $b=4$.

If $c=3$ then in adding the hundreds digits we must have a carry to the thousands digit which is possible only when $a=9$. But by the ones digit we should have $c=5$. The contradiction shows that this case is not possible.