In the decimal system, the reverse of the $(n+1)$-digit number $A=\overline{a_{n} a_{n-1} \ldots a_{1} a_{0}}$ is the number $A^{*}=\overline{a_{0} a_{1} \ldots a_{n}}$. (Thus, the reverse of 759 is 957, and the reverse of 980 is 89.)

Find the four-digit numbers that "reverse" when multiplied by 9, i.e., for which $9 A=A^{*}$.
Clearly, $1000 \leq A \leq 1111$, since $A^{*}$ is at most a four-digit number, and if $A > 1111$, then $9A$ would already be a five-digit number. Therefore,

$$
9000 \leq 9A \leq 9999
$$

Thus, $A^{*}$ is exactly a four-digit number, its first digit is 9, so the last digit of $A$ is 9. Since $A^{*}$ is divisible by 9, $A$ is also divisible by 9, which means the sum of its digits is divisible by 9. This implies that the sum of the first three digits of $A$ is also divisible by 9. Since the first digit of $A$ is 1, and the second digit is 0 or 1, the third digit is 8 or 7. However, in the latter case, $A$ would be 1179, which is too large; the only possibility is $A = 1089$, and for this, the condition is indeed satisfied, since $9 \cdot 1089 = 9801$.

Thus, the problem has only one solution, which is 1089.

Matolcsi Máté (Bp., Fazekas M. Gyak. Gimn., I. o. t.)

Note. $A$ is a square number, so $A^{*}$ is also a square number.