1. Find all four-digit numbers that are written with four consecutive digits, in any order, such that their product with $\frac{2}{3}$ is a number written with the same digits.
Solution. Let the number be $x$ and let $y=\frac{2}{3} x$. From the last equality, we have $3 y=2 x$, so $3 \mid x$ and $2 \mid y$. From the divisibility rule by 3, it follows that the sum of the digits of the number $x$ is divisible by 3. However, the number $y$ is written with the same digits as the number $x$, so $3 \mid y$.

From what has been stated so far, we have $2|y, 3| y$ and $\operatorname{NZD}(2,3)=1$, so $y=6 k$, which means $x=\frac{3}{2} y=\frac{3}{2} \cdot 6 k=9 k$. Therefore, $9 \mid x$ and from the divisibility rule by 9, we get that the sum of the digits of the number $x$ is divisible by 9. Clearly, the sum of the digits of the number $y$ is also divisible by 9. However, the numbers $x$ and $y$ are written with four consecutive digits $n, n+1, n+2, n+3$, so their sum is $4 n+6, n \in \mathbb{N}$ and $n \leq 6$. Thus,

$$
9 \mid 4 n+6, n \in \mathbb{N} \text { and } n \leq 6
$$

Condition (1) is satisfied only for $n=3$, from which it follows that the digits with which the numbers $x$ and $y$ are written are $3,4,5$ and 6.

From what has been stated so far, we have $3456 \leq x \leq 6543$ and since $y=\frac{2}{3} x$ we get $2304 \leq y \leq 4362$. However, the number $y$ is written with the digits $3,4,5$ and 6 and is divisible by 2, so possible solutions for $y$ are the numbers

$3456,3546,3564,3654,4356$ and 4536,

to which the following solutions for the number $x$ correspond

$$
5184,5319,5346,5481,6534 \text { and } 6804 .
$$

However, $x$ is written with the digits 3,4,5 and 6, so the only solutions are 5346 and 6534.