1. A certain 4-digit number was added to the number written with the same digits but in reverse order, and the result was 4983. What numbers were added
Answer: 1992 and 2991.

Solution: Let's write the condition as $\overline{a b c d}+\overline{d c b a}=1001(a+d)+110(b+$ $c)=4983$. Notice that $a+d$ must end in 3, so $a+d=3$ or $a+d=13$. The second case is impossible, since then $1001(a+d)+110(b+c)>13000$. Therefore, $a+d=3$, substituting into the original equation, we get $3003+110(b+c)=4983$, from which $b+c=18$, so $b=c=9$.