2. Find all digits $n$ and 2018-digit natural numbers $x, x=\overline{a_{2017} \ldots a_{2} a_{1} a_{0}}$, such that

$$
n \cdot x=\overline{\left(a_{2017}+n\right) \ldots\left(a_{2}+n\right)\left(a_{1}+n\right)\left(a_{0}+n\right)}
$$
2. Let $y=\overline{\left(a_{2017}+n\right) \ldots\left(a_{2}+n\right)\left(a_{1}+n\right)\left(a_{0}+n\right)}$. Clearly, the digit $n$ is neither 0 nor 1.

We will prove that $a_{0}$ is an even number. Assume the contrary. Then $x$ is an odd number, and since $y=n \cdot x$, $y$ is even if and only if $n$ is even. However, $a_{0}+n$ has a different parity from $n$, which is a contradiction. Therefore, $a_{0}$ is an even number, and both $x$ and $y$ are even numbers. Hence, $a_{0}+n$ is also even, so $n$ is even.

Also, $a_{0} \neq 0$, because if it were, the units digit of $y$ would be 0, which is impossible. Therefore, $a_{0} \geqslant 1$, and since $a_{0}$ is even, it follows that $a_{0} \geqslant 2$. From $a_{0}+n \leqslant 9$ we get $n \leqslant 7$, and since $n$ is even, it follows that $n \leqslant 6$. Thus, $n \in\{2,4,6\}$.

For $n=6$, from $a_{0}+n \leqslant 9$ we get $a_{0}=2$ and $a_{0}+n=8$. However, the units digit of the number $6 \cdot \overline{a_{2017} \ldots a_{2} a_{1} 2}$ is 2, not 8.

For $n=4$, from $a_{0}+n \leqslant 9$ we get $a_{0}=2$ or $a_{0}=4$, and $a_{0}+n=6$ or $a_{0}+n=8$, respectively. However, the units digit of the number $4 \cdot \overline{a_{2017} \ldots a_{2} a_{1} 2}$ is 8, and the units digit of the number $4 \cdot \overline{a_{2017} \ldots a_{2} a_{1} 4}$ is 6, so this case is also impossible.

Therefore, the only remaining case is $n=2$, and then $a_{0} \in\{2,4,6\}$. By comparing the units digits of the number $2 \cdot x$ and $y$, we conclude $a_{0}=2$. Now from

$$
2 \cdot \overline{a_{2017} \ldots a_{2} a_{1} 2}=\overline{\left(a_{2017}+n\right) \ldots\left(a_{2}+n\right)\left(a_{1}+n\right) 4}
$$

it follows that

$$
2 \cdot \overline{a_{2017} \ldots a_{2} a_{1}}=\overline{\left(a_{2017}+n\right) \ldots\left(a_{2}+n\right)\left(a_{1}+n\right)}
$$

Further, in the same way, we get $a_{1}=2, a_{2}=2$, and so on, up to $a_{2017}=2$. Therefore, the only solution is $n=2$ and $x=\underbrace{222 \ldots 222}_{2018 \text { times }}$.

![](https://cdn.mathpix.com/cropped/2024_06_04_841f52b4ffb037ad799bg-41.jpg?height=476&width=505&top_left_y=1002&top_left_x=187)

Ok 2018 1A 3