In the following multiplication schemes, each letter represents a digit, with the only constraint being that the same letter always represents the same digit. Determine the value of the letters so that the multiplications are correct.

| $\frac{A B}{} \cdot C D$ | $\cdot A B$ | $C G \cdot K L$ |
| :--- | :--- | :--- |
| $E F G$ | $E F G$ | AMC |
| $\frac{E F G}{H I G G}$ | $\frac{E F G}{H I G G}$ | $\frac{A M C}{N P C C}$ |
In all three schemes, the two-by-two partial products are equal, so the digits of the multipliers are equal: $C=D, A=B, K=L$.

In the addition of all three schemes, the digit in the tens place is equal to one of the addend digits. There cannot be a carry yet, so the value of the other addend digit is 0, that is, $F=M=0$. Thus, there is no carry from the tens column to the hundreds column anywhere, and from the hundreds to the thousands, so $I=E=H$, and $P=A=N$. We only need to determine the values of the letters $A, C, E, G, K$.

The multiplicand in all three schemes is divisible by 11: $\overline{A B}=11 A, \overline{C D}=11 C, \overline{G G}=11 G$, so for the partial products we have

$$
\overline{E 0 G}=11 A C, \quad \overline{A 0 C}=11 G K
$$

or

$$
\begin{gathered}
100 E+G=99 E+(E+G)=11 A C, \quad \text { and } \\
99 A+(A+C)=11 G K, \quad \text { from which } \\
E+G=11(A C-9 E), \quad A+C=11(G K-9 A)
\end{gathered}
$$

thus $A+C$ and $E+G$ are divisible by 11. On the other hand, $A, C, E, G$ are none of them 0, as each appears as a leading digit, and none is greater than 9, so

$$
E+G=11, \quad \text { and } \quad A+C=11
$$

Therefore, from (1)

$$
E=\frac{A C-1}{9}
$$

and

$$
K=\frac{9 A+1}{G}
$$

Accordingly, neither $A$ nor $C$ can be 9, because then $E$ could not be an integer. But $A$ and $C$ cannot be divisible by 3 either, otherwise $A C$ would be a multiple of 3, so it could not be 1 greater than $9 E$. Thus, according to (2), only the 4, 7 digit pair can be used for $A$ and $C$. Then from (3) in any case $E=3$, from (2) $G=8$, and from (4)

$$
K=\frac{9 A+1}{8}=A+\frac{A+1}{8}
$$

Thus only $A=7$ and $C=4$ are possible, and furthermore $K=8$.

According to all this, the three schemes can only represent the multiplications $77 \cdot 44$, $44 \cdot 77$ and $88 \cdot 88$. These indeed meet the conditions.

Zoltán Kunszt (Pápa, Türr I. Gym. III. o. t)

Remarks. 1. There is another way to further determine the value pair of $A, C$. According to (2) and (3), $A$ and $C$ are the roots of the quadratic equation

$$
x^{2}-11 x+9 E+1=0
$$

so

$$
x_{1,2}=A, C=\frac{11 \pm \sqrt{117-36 E}}{2}
$$

The discriminant cannot be negative: $117-36 E \geq 0$, so $E1$. Now, with $E=2$, $x$ is not rational, but with $E=3$, $A, C=4,7$.

Károly Somogyi (Bonyhád, Petőfi S. Gym. III. o. t.)

2. $111=A+C$. Thus, the value of $A C$ can only be 28.