Problem 4. Determine the non-zero natural numbers $A$ and $B$, which have the same number of digits, knowing that

$$
2 \cdot A \cdot B = \overline{A B}
$$

where $\overline{A B}$ is the number obtained by writing the digits of $B$ after the digits of $A$.
Solution Let $n$ be the number of digits of $A$ and $B$. From the relation in the statement we have $(2 A-1) B=10^{n} A$,

![](https://cdn.mathpix.com/cropped/2024_06_07_633976caf00b87b15028g-3.jpg?height=57&width=1797&top_left_y=1554&top_left_x=175)

Since $(2 A-1, A)=1$ and $(2,2 A-1)=1$, it follows that $2 A-1 \mid 5^{n}$, so $2 A-1 \leq 5^{n} \ldots \ldots \ldots .$. . 2 p

Since $A$ has $n$ digits, it follows that $A \geq 10^{n-1}$, so $2 \cdot 10^{n-1}-1 \leq 2 A-1 \leq 5^{n}$. We obtain

$$
2 \cdot 2^{n-1} \cdot 5^{n-1} \leq 5 \cdot 5^{n-1}+1 \leq 6 \cdot 5^{n-1}
$$

thus $2^{n} \leq 6$, from which $n \in\{1,2\}$...

For $n=1$, from $2 A-1 \mid 5$, it follows that $2 A-1 \in\{1,5\}$, so $A \in\{1,3\}$. If $A=1$, we have $B=10$, which does not work since $B$ has one digit. If $A=3$, we have $B=6 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$............................

For $n=2$, from $2 A-1 \mid 25$, it follows that $2 A-1 \in\{1,5,25\}$, so $A \in\{1,3,13\}$; but $A$ has two digits, so $A=13$ and $25 B=1300$, so $B=52$. $1 p$