## Task 4 - 120734

When class 7a entered the math classroom, a multiplication problem was written on the blackboard. However, someone had smudged the digits so much that only four "ones" remained legible, and of the illegible digits, only their positions could still be recognized.

The image on the blackboard looked like this: (The illegible digits are indicated here by the letters $a, b, c, \ldots$. Thus, different letters can also represent the same digit, possibly even the digit 1 again.)

|  |  | 1 | a | b | $\cdot$ | c | d |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
|  | $e$ | $f$ | g | 1 |  |  |  |
| $\mathrm{~h}$ | $\mathrm{i}$ | $\mathrm{j}$ | 1 |  |  |  |  |
| $\mathrm{k}$ | $\mathrm{m}$ | $\mathrm{n}$ | 1 | p |  |  |  |

Some students immediately tried to determine the missing digits, and after a short while, Bernd called out:

"I know exactly what the two factors were!"

But Gerd countered him:

"It cannot be determined uniquely what the two factors were."

Determine whether Bernd or Gerd was right! In each case, provide all solutions (realizations) of the multiplication schema!
Clearly, $p=1$ and $g=0$. Since a four-digit number is present in both the second and third lines, $d \neq 1$ and $c \neq 1$. The product $b \cdot d$ ends in the same digit as the product $b \cdot c$, which is 1. There are now exactly the following three cases:

1. Case: Let $b=9, c=d=9$.

This implies $b \cdot d=81$. Since $g=0$, the product $a \cdot d$, which is $a \cdot 9$, must end in the digit 2 (since $2+8=10$). This is true only for $a=8$. Thus, the first factor is 189, and the second is 99. Indeed, the calculation $189 \cdot 99$ results in a multiplication table that matches the problem statement.

2. Case: Let $b=7, c=d=3$.

Since the first factor is less than 200, its triple is less than 600, which is a three-digit number, contradicting the second line. Therefore, this case does not lead to a solution.

3. Case: Let $b=3, c=d=7$.

This implies $b \cdot d=21$. The product $a \cdot d$, which is $a \cdot 7$, must then end in the digit 8 (since $2+8=10$). This is true only for $a=4$. Thus, the first factor is 143, and the second is 77.

Indeed, the calculation $143 \cdot 77$ results in a multiplication table that matches the problem statement.

Thus, the multiplication schema has exactly two, and specifically the following two, realizations:

|  |  | 1 | 8 | 9 | $\cdot$ | 9 | 9 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
|  | 1 | 7 | 0 | 1 |  |  |  |
| 1 | 7 | 0 | 1 |  |  |  |  |
| 1 | 8 | 7 | 1 | 1 |  |  |  |


|  |  | 1 | 4 | 3 | $\cdot$ | 7 | 7 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
|  | 1 | 0 | 0 | 1 |  |  |  |
| 1 | 0 | 0 | 1 |  |  |  |  |
| 1 | 1 | 0 | 1 | 1 |  |  |  |

Thus, Gerd was right, while Bernd was not.