Problem 1960

Since the numbers $n$ and $k$ are positive integers of the same length, the quotient $\frac{n}{k}$ must be a single-digit positive number, because multiplying by a multi-digit number increases the number of digits. The quotient $1$ is obviously possible (e.g. $\frac{69}{69} = 1$). We show that no other quotient is possible.

* If the first digit of the number $k$ is $1$, then the last digit of the number $n$ is $1$. The quotient $\frac{n}{k}$ cannot be $2$, $4$, $5$, $6$, or $8$, because the multiples of these numbers cannot end with the digit $1$. If the quotient $\frac{n}{k}$ were $3$ or $7$, then the last digit of $k$ should be $7$ or $3$, respectively. However, these digits cannot occur in a twisting. If the quotient $\frac{n}{k}$ were $9$, then the last digit of the number $k$ should be $9$ and the first digit of the number $n$ should therefore be $6$. However, since $n = 9k$, the number $n$ can only start with the digit $9$. Therefore, the only possibility is $\frac{n}{k} = 1$.

* If the first digit of the number $k$ is $2$, then the last digit of the number $n$ is $5$. The quotient $\frac{n}{k}$ cannot be $5$, $6$, $7$, $8$, or $9$, because in these cases the number $n$ would have more digits than the number $k$. The quotient $\frac{n}{k}$ cannot be $2$ or $4$, because the multiples of these numbers cannot end with the digit $5$. If the quotient $\frac{n}{k}$ were $3$, then the last digit of the number $k$ should also be $5$ and the first digit of the number $n$ should therefore be $2$. However, since $n = 3k$, the number $n$ can only start with the digits $6$, $7$, and $8$. So again, the only possibility is $\frac{n}{k} = 1$.

* The first digit of the number $k$ cannot be $3$ or $4$, because these numbers do not occur in a twisting.

* If the first digit of the number $k$ is $5$, $6$, $7$, $8$, or $9$, then $\frac{n}{k} \ge 2$ is not possible, because then there would be more digits in the number $n$ than in the number $k$. Thus, the only possibility is $\frac{n}{k} = 1$.