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

$$
\begin{array}{rlrl}
A B \cdot C A & =D E B \\
+\quad+ & = & : \\
F C-D G & = & D \\
\hline E G+H J & =A A G
\end{array}
$$
The last digit of the 3rd sum can only be $G$ if $J=0$. On the other hand, since the sum is less than 200, $A=1$. Thus, from the addition of the unit digits in the 2nd column, $G=9$. Therefore, on one hand, in the subtraction trial of the 2nd row, the sum of the unit digits $D+G=D+9$ can only be $C+10$, so $C=D-1$, and consequently, from the tens place digits, $F=D+1$. On the other hand, in the addition of the unit digits in the 1st column, there cannot be a carry, because with $G=9$, $B+C \leqq 8+7=15$, so $B+C=9, B=9-C$, and furthermore, from the tens place digits, $E=F+A=F+1=D+2$. In the tens place of the 2nd column, considering the carried over digit, $C+D+1=H$, so substituting $C$, $H=2D$. Thus, from the tens and hundreds place digits of the 3rd row, $E+H=(D+2)+2D=3D+2=11$, so $D=3$. From this, based on the previous findings, $C=2, F=4, E=5, B=7$, and $H=6$.

$$
\begin{gathered}
17 \cdot 21=357 \\
+\quad+\quad: \\
42-39=3 \\
\hline 59+60=119
\end{gathered}
$$

With this, we have assigned a definite numerical value to every letter, and if the problem has a solution, it can only be this value system. Substitution shows that with this value system, all operations yield the correct result.

Zsigmond Mocskónyi (Sopron, Forestry Technical College, 1st year apprentice)

Remarks. $A=1$ also follows from the division of the hundreds place digit in the 3rd column. - However, it is incorrect to assume $A=1$ from the multiplication in the 1st row, because the product $A \cdot B$ can end in $B$ even if $A=6$ and $B$ is even, or if $B=5$ and $A$ is odd.

If we already know that $A=1, G=9$, and $E=D+2$, then after determining the tens place digit $A$ of the quotient in the 3rd column, the remainder is 2 tens, so the value of the quotient $(20+B): D$ is $G=9$. From this, $D=3$ and $B=7$.

Several people found that only $D=3$ is possible after many unsuccessful attempts based on assumptions about the value of $D$. The number of these attempts can be reduced to two by considering that $AAG=119$ and the first row; from the first row, it is clear that $CC=5$ and $D \neq G$, so only $D=6$ is possible, but this leads to a contradiction in the 2nd column.