1. A certain three-digit number was added to the number written with the same digits but in reverse order, and the result was 1777. What numbers were added

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Answer: 839 and 938.

Solution: Let the number be $\overline{a b c}=100 a+10 b+c$, then $\overline{a b c}+\overline{c b a}=$ $101 a+20 b+101 c=1777$. The sum $a+c$ must end in 7, i.e., $a+c=7$ or $a+c=17$. The first case is impossible, since then $101 a+20 b+101 c \leqslant 7 \cdot 101+9 \cdot 20=777+180<1777$. Therefore, $a+c=17$, which is only possible if $a=8, c=9$ or vice versa. Substituting into the equation, we get $707+20 b=1777$, from which $b=3$.