2. When a five-digit number is multiplied by 9, the result is a number composed of the same digits but in reverse order. Find the original number.
Answer: 10989.

Let $\overline{a b c d e}$ be the original number. The condition is written as the equation $9 \cdot \overline{a b c d e}=\overline{e d c b a}$. Note that $a=1$, because if $a \geqslant 2$, then $9 \cdot \overline{a b c d e} \geqslant 9 \cdot 20000>100000>\overline{e d c b a}$.

We have $9 \cdot \overline{1 b c d e}=\overline{e d c b 1}$. From this, the last digit $e=9$ is uniquely determined.

Working with $9 \cdot \overline{1 b c d 9}=\overline{9 d c b 1}$. Note that if $b \geqslant 2$, then $9 \cdot \overline{1 b c d 9}>$ $9 \cdot 12000>100000>\overline{e d c b a}$. Therefore, $b=0$ or $b=1$. Let's consider these two cases:

If $b=0$, the equation reduces to $9 \cdot \overline{10 c d 9}=\overline{9 d c 01}$. Considering the second-to-last digit, we conclude that $d=8$. By trying the possible values of the digit $c$, we find the unique solution $\overline{a b c d e}=10989$.

If $b=1$, the equation is rewritten as $9 \cdot \overline{11 c d 9}=\overline{9 d c 11}$. Analyzing the second-to-last digit, we get $d=7$. But $9 \cdot \overline{11 c 79}>9 \cdot 11000=99000>\overline{97 c 11}$, which is a contradiction. Therefore, there are no solutions in the case $b=1$.

The only number that satisfies the condition of the problem is 10989.