1. Find the largest five-digit natural number that is divisible by 101 and reads the same forwards and backwards.
1. Any of the considered five-digit numbers has a decimal representation of the form $\overline{a b c b a}$. By expanding and rearranging it, we obtain the equality

$$
\overline{a b c b a}=10001 a+1010 b+100 c=101(99 a+10 b+c)+2 a-c .
$$

From this, it follows that the number in question is divisible by 101 if and only if $2 a-c=0$ (for any digits $a, c$, it certainly holds that $|2 a-c|<101$). From the equality $2 a=c$, it follows that $a \leqq 4$, and since we are looking for the largest such number, we choose its first digit $a=4$, which corresponds to the digit $c=8$. Since the digit $b$ does not affect the divisibility by 101, we choose it to be as large as possible: $b=9$. Therefore, the sought number is 49894.

For a complete solution, award 6 points.