## Task 1 - 130821

In the following task, the letters $a, b, c$ and the symbol $*$ are to be replaced by one of the digits 0 to 9 each, so that a correctly solved and conventionally written multiplication problem results.

Identical letters represent identical, different letters represent different digits. No equality or difference requirements are imposed on the digits to be set for the symbols *.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0661.jpg?height=206&width=469&top_left_y=474&top_left_x=1276)
Assuming there is an entry of the required type. Then it follows:

The product of $a b c$ and $a$ is three-digit, while the product of $a b c$ and $b$ is four-digit. Therefore, $a < b$. If $a \geq 3$, then $b \geq 4$, and thus the product of $abc$ and $a$ would be four-digit, contradicting the problem.

The product of $a b c$ and $a$ ends in $a$. If $a = 1$, then this product would end in $c$, contradicting $a \neq c$. Therefore, and because $a$ as the leading digit of $a b c$ cannot be 0, $a = 2$.

Since the double of the number $a b c$ ends in 2, the double of $c$ must also end in 2. This is true only for $c = 1$ or $c = 6$. Since the product of $a b c$ and $c$ is four-digit, $c \neq 1$. Therefore, $c = 6$.

Since $246 \cdot 4 = 984$ is three-digit, but the product of $a b c$ and $b$ should be four-digit, $b \neq 4$. Among the remaining possibilities for $b$ (1, 3, 5, 7, 8, 9), only the number 8 satisfies the condition that the product of the number ending in 6 with $b$ ends in $b$. Therefore, $b = 8$.

Thus, only the entry

$$
\begin{array}{ccccccc}
2 & 8 & 6 & \cdot & 2 & 8 & 6 \\
\hline & 2 & 2 & 8 & 8 & & \\
& & & 5 & 7 & 2 & \\
& & & 1 & 7 & 1 & 6 \\
\hline & 2 & 3 & 6 & 2 & 3 & 6
\end{array}
$$

can meet the requirements. Since it represents a correctly solved multiplication problem and since $a = 2$, $b = 8$, $c = 6$ are pairwise distinct, it satisfies the conditions of the problem.