9.1. A four-digit number $n$ was doubled and 1000 was added to the result. As a result, a four-digit number was obtained, written with the same digits as $n$, but in reverse order. Find all possible values of $n$.
Answer: 2996.

Solution. Let $a, b, c, d$ be the number of thousands, hundreds, tens, and units in the number $n$ respectively. Write the condition in the form:

| $a$ | $b$ | $c$ | $d$ |
| ---: | ---: | ---: | ---: |
| $a$ | $b$ | $c$ | $d$ |
| $+\quad 1$ | 0 | 0 | 0 |
| $d$ | $c$ | $b$ | $a$ |

Since $a$ is the last digit of $2d$, $a$ is even. Since $2a+1 \leq d \leq 9$, it follows that $a \leq 4$. Therefore, either $a=2$ or $a=4$.

If $a=4$, then from $2a+1 \leq d \leq 9$ it follows that $d=9$. But then the last digit of the number $2d$ is 8, and $a \neq 8$.

If $a=2$, then from $2a+1 \leq b \quad c \quad d$ it follows that $d \geq 5$. And since the last digit of $2d$ is 2, then $d=6$. Thus, the equation

![](https://cdn.mathpix.com/cropped/2024_05_06_e21db4466835fed407acg-1.jpg?height=140&width=294&top_left_y=1780&top_left_x=1018)
reduces to the form:

Here, either $2c+1=b$ and $2b=10+c$, or $2c+1=10+b$ and $2b+1=10+c$. In the first case, the system has no integer solutions, in the second case we get $b=c=9$.