Which is the three-digit number that is 1 less than twice the number obtained by swapping its extreme digits?
I. solution. Denoting the digits of the sought number by $x, z, y$,

$$
\begin{gathered}
\overline{x z y}+1=2 \cdot \overline{y z x}, \quad \text { i.e., } \\
100 x+10 z+y+1=2(100 y+10 z+x)
\end{gathered}
$$

From this, expressing $z$,

$$
z=10(x-2 y)-\frac{2 x-y-1}{10}
$$

Since $1 \leq y \leq 9$ and due to the subtraction of 1 unit not changing the tens place, $z=9$. Furthermore, the sum of the extreme digits of the difference is also 9, so $x+y=10$. Finally, since $y$ is odd and $y<x$, we only need to test the numbers 793 and 991.

Lehel Jenö (Budapest, Apáczai Csere J. practice g. III. class)

2. Noting the similarity of our problem to the 716th exercise, it is easy to see that for any $k \geq 2$, there is a $k$-digit (natural) number which is 1 less than twice the number obtained by swapping its extreme digits, specifically the number $M=7999 \ldots 93$. This number is formed by inserting $k-2$ nines between the digits of 73. Indeed, in another form, $M=8 \cdot 10^{k-1}-7$, and the number obtained by swapping the digits is $N=3999 \ldots 97=4 \cdot 10^{k-1}-3$, and thus $2 N-1=M$.

Huber Tibor (Budapest, Kossuth L. mechanical engineering t. IV. class)

3. If we allow a decimal point in the sought number, we are looking for solutions of the form $\pm \overline{x y, z} \pm \overline{x, y z}$ and $\pm 0, x y z$. The number $-6.92$ fits the second type.

Kotsis Domokos (Budapest, József A. g. III. class) [^0]

[^0]:    ${ }^{1}$ When the number obtained by writing the digits of a 3-digit (natural) number in reverse order is subtracted from the number, and the extreme digits are different, the extreme digit of the difference can be determined from the difference.

    ${ }^{2}$ See K. M. L. 24 (1962) 121.