How many natural numbers are there whose representations in both base 9 and base 11 are three-digit numbers? Determine among them those numbers whose representations in these two bases consist of the same digits but in reverse order.
a) The three-digit natural numbers in the base-9 number system range from $9^{2}=81$ to $9^{3}-1=728$, while in the base-11 number system, they range from $11^{2}=121$ to $11^{3}-1=1330$. The largest number that is three digits in both systems is 728, and the smallest is 121, so the number of such numbers is $728-(121-1)=608$.

b) Let the sought number have the form $\overline{A B C}_{9}=81 A+9 B+C$ in the base-9 number system, and thus $\overline{C B A}_{11}=121 C+11 B+A$ in the base-11 number system, with the digits satisfying $1 \leq A, C \leq 8$ and $0 \leq B \leq 8$. From the equality of the two expressions,

$$
B=40 A-60 C=20(2 A-3 C)=20 k,
$$

where $k$ is an integer. This is only possible if $B=0$ and $k=0$, i.e., $2 A=3 C$, which means $A$ must be divisible by 3, so $A=3$ or 6, and consequently $C=2$ or 4. The following two numbers satisfy the condition:

$$
302_{9}=203_{11}=245_{10}, \quad 604_{9}=406_{11}=490_{10} .
$$

Péter Rimóczi (Debrecen, Kodály Z. Music School II. Grade)