Which is the two-digit number that is 1 less than twice the number obtained by swapping its digits? (H)
I. solution. Let the sought number be $\overline{x y}=10 x+y$, then

$$
\overline{x y}+1=2 \cdot \overline{y x} \quad \text { that is } \quad 10 x+y+1=20 y+2 x
$$

It is clear from this that $x>y$. The right side is even, so the left side must also be, hence $y$ is odd. Furthermore, $y$ is less than 5, otherwise $x$ would be at least 6, the right side would be at least 112, and thus it could not be just 1 more than a two-digit number. Thus, the value of $y$ is 1 or 3. With $y=1$, $x$ is not an integer, but with $y=3$, $x=7$. The number 73 indeed meets the requirement, and there are no other such two-digit numbers.

Jáky Géza (Pécs, Széchenyi I. g. I. o. t.)

Note. The conclusions about $y$ can also be read from the expression obtained by rearranging (1)

$$
x=2 y+\frac{3 y-1}{8}
$$

Indeed, $x, y$ are integers, so $3 y-1$ is even, hence $y$ is odd; also, $2 y<x \leq 9$, so $y \leq 4$.

II. solution. Writing the requirement as

$$
\overline{x y}+1=\overline{y x}+\overline{y x}
$$

from the sum of the units place digits

$$
y+1=x+x, \quad \text { or } \quad y+1+10=x+x
$$

depending on whether there is no carry or there is a carry to the tens place. The first possibility yields $y \geq x$, which is impossible. The second gives $y=2 x-11$, and from the tens place digits, together with the carried remainder,

$$
x=2 y+1
$$

From these two equations, $x=7, y=3$.

Friss Ilona (Budapest, Radnóti M. gyak. g. III. o.t.)