11.2. Let $n$ be a natural number not ending in 0, and $R(n)$ be the four-digit number obtained from $n$ by reversing the order of its digits, for example $R(3257)=7523$: Find all natural four-digit numbers $n$ such that $R(n)=4n+3$.
Answer: 1997.

Solution. Consider the decimal representation of the original four-digit number $n=\overline{a b c d}$, then $R(n)=\overline{d c b a}=4 n+3$ is also a four-digit number. Therefore, $4 a \leq 9$, so $a=1$ or $a=2$. Moreover, the number $R(n)=\overline{d c b a}=4 n+3$ is odd and ends in $a$, hence $a=1$. If the number $4 n+3$ ends in 1, then $n$ must end in 2 or 7, that is, $d=2$ or $d=7$. The first is impossible, since $d \geq 4 a \geq 4$, so $d=7$. Substituting the found values into the equation from the condition: $R(n)=4 \cdot \overline{a b c d}+3=4031+400 b+40 c=7001+100 c+10 b$, from which $13 b-2 c=99$. From this, $b \geq \frac{99}{13}=7 \frac{8}{13}$, that is, $b=8$ or $b=9$. When $b=8$, the left side of the equation $13 b-2 c=99$ is even, which is impossible, so $b=9$ and $c=9$.

Grading criteria. Answer: 2 points. Proof that there are no other answers: 5 points.