In the standing equalities, each letter represents a digit, different letters represent different digits. Below the line, the result of the operation indicated in the column stands. Determine the value of the letters so that all six operations yield the correct result.

$$
\begin{gathered}
a b c-d c=a e f \\
: \quad+\quad- \\
g \cdot h=e i \\
\hline j j+g d=b c
\end{gathered}
$$
From the first row, $f=0$, as the difference of the matching "1" place value digits. From the 3rd column, $a=1$, because as a trial subtraction, the sum of the two-digit numbers $ei$ and $bc$ is a three-digit number less than 200: furthermore, due to $f=0$, $c+i=10$ and $1+b+e=10+e$, so $b=9$. - Thus, the dividend in the first column is on the one hand at least 192 and at most 198, and on the other hand, according to the quotient $\overline{jj}$, a multiple of 11. Among the mentioned numbers, only $198=2 \cdot 3^{2} \cdot 11$ fits, so $c=8$. Thus, from the 3rd column, $i=2$. Again from the first column, $g$ is a one-digit divisor of 198, different from $1=a$, $2=i$, and $9=b$, so $g=3$ or $g=6$. However, $g=3$ leads to $d=g-1=i$ through the tens digits of the 2nd column, so $g=6, d=5$, and from the first column, $j=3$. From the 2nd column, $h=65-58=7$, and from the 2nd row, $e=4$.

$$
\begin{gathered}
198-58=140 \\
\vdots \quad++ \\
6 \cdot 7=42 \\
\hline 33+65=98
\end{gathered}
$$

With these letters written in place of the variables, the partially used requirement of the first row and the so far ignored requirement of the 3rd row are also satisfied, the solution is correct.

Judit Barabás (Budapest, Apáczai Csere J. Practical Grammar School I. Class)