In the following equalities, each letter represents a digit, and different letters represent different digits. Under the line, in each column, the result of the operation indicated vertically is given. Determine the value of the letters so that both the horizontal and vertical operations appear with the correct result.

$$
\begin{aligned}
\overline{a b}-\overline{c d} & =\overline{e d} \\
: \quad- & =- \\
e \cdot f & =\overline{d c} \\
\hline \overline{g d}+\overline{g b} & =\overline{h c}
\end{aligned}
$$
We denote the operations of consecutive rows and columns as I, II, III, and IV, V, VI, respectively; and we refer to our further findings with the following consecutive numbers.

1. Except for $b$, every letter appears as the first digit of some number, so their value cannot be 0.
2. The value of $e$ and $f$ cannot be 1 either, because according to II., their product is a two-digit number.
3. From I., $b$ and from VI., $d$ are even, because 1 is the unit digit of $2d$ and $2c$, respectively. Therefore, from III., $c$ and from V., $f$ are also even, as the unit digit of the sum and difference of $d$ and $b$. (The evenness of $c$ also follows from $f$ according to II.)
4. Now, $b$ cannot be 0 either, because from $b=0$ and I., either $d=0$ or $d=5$ would follow, which we have already excluded in 1. and 3., respectively. Thus, the digit 0 does not appear, $b, c, d, f$ are even digits, and consequently, $a, c, g, h$ are odd.
5. From the tens digits in III., $g<5$, and considering its oddness, the value of $g$ is 1 or 3. Then, from V., the value of $c$ is 2 or 4.
6. From II., $e \cdot f \leqq 9 \cdot 8=72$, so $d \neq 8$. From VI. and 4., $c \neq 4$. From 5., $c=2$, from V., $g=1$, and from III. and 4., $h=3$. - These value determinations mean that the values of $c, g, h$ cannot be anything other than 2, 1, and 3, respectively; or these values are part of the solution, or there is no solution.
7. From $c=2$ and VI., $d=4$, from I., $b=8$, from these and 4., $f=6$, and from II., $e=42: 6=7$, and finally from IV., $\overline{a b}=e \cdot \overline{g d}=7 \cdot 14=98$ gives $a=9$.

$98-24=74 \quad$ It only remains to check whether

: $\quad-\quad-\quad$ substituting the only possible value for each

7 . $6=42$ digit, all operations are correct. The affirmative

$\overline{14+18=32}$ answer shows that the problem is uniquely solvable.

Peter Grünfeld (Bp. IX., József A. Mechanical Engineering Technology II. o. t.)