7.4. Does there exist a six-digit natural number that, when multiplied by 9, is written with the same digits but in reverse order?
Answer: it exists. Solution. Let $\overline{a b c d e f}$ be the desired number, i.e., $\overline{a b c d e f} \cdot 9=\overline{f e d c b a}$. Then it is obvious that $a=1, b=0$ (otherwise, multiplying by 9 would result in a seven-digit number). Therefore, $f=9$, and the second-to-last digit $e=8$ (which follows from column multiplication). Then the third digit $c$ can be 8 or 9. But if $c=8$, then $d=1$, since the sum of the digits is divisible by 9, but the number 108189 does not fit upon verification. If, however, $c=9$, then $d=9$ and the number 109989 is the only one that satisfies the condition, and it fits upon verification.