Sémánk egy, a nyolc-alapú számrendszerben végrehajtott szorzást (négyzetreemelést) mutat. Egyező betúk egyenlő számjegyeket jelentenek, különböző betúk különböző számjegyeket, a pontok pedig valamilyen számjegyeket. Azt is tudjuk, hogy az alapban és a négyzetében a rendszer mindegyik számjegye pontosan egyszer lép föl.

Keressünk ilyen $A B C$ számot.

![](https://cdn.mathpix.com/cropped/2024_05_02_0e34abe4cc15da134b76g-1.jpg?height=257&width=290&top_left_y=300&top_left_x=907)

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Here is the translation:

This diagram shows a multiplication (squaring) in the octal number system. Matching letters represent the same digits, different letters represent different digits, and the dots represent some digits. We also know that in both the base and its square, each digit of the system appears exactly once.

Find such an $A B C$ number.

![](https://cdn.mathpix.com/cropped/2024_05_02_0e34abe4cc15da134b76g-1.jpg?height=257&width=290&top_left_y=300&top_left_x=907)
As $A B C \cdot A$ partial product starts with $B$ and is three-digit. Therefore, on one hand $A \neq 1$, and on the other hand $A < 8$, otherwise the product would already be four-digit, so the value of $A$ can only be 2.

Thus, $200 \leqq A B C \leqq 277$, so $400 \leqq A \cdot A B C=B \cdot D \leqq 576$, hence the value of $B$ is 4 or 5 (we also understand these calculations in the octal number system). Using this again, $240 \leqq A B C \leqq 277$ and $500 \leqq 2 \cdot A B C=B \cdot D \leqq 576$, so the value of $B$ can only be 5.

Now we can estimate the second partial product as follows:

$$
B \cdot A B C=5 \cdot \overline{25 C}B$, because the leading digit of the $C \cdot A B C$ partial product is 2; therefore, the value of $C$ can only be 6 or 7.

Finally, the last digit of the second partial product is $C$, so the product $B \cdot C$ ends in $C$, the difference $B \cdot C-C=4 C$ is divisible by 8, $C$ is even, its value can only be 6.

According to the above, the only possible multiplication is $256 \times 256$, but this is only a solution to the problem if the further, not yet used, conditions are also met. In this multiplication, the first and third partial products end in 4, which is suitable for $D$ because 4 has not appeared in our previous considerations, and 256 and $256^{2}=73024$ use each digit of the octal system exactly once. Therefore, the number $A B C$ can only be 256, and only 256 is suitable.

| $256 \times 256$ |
| :--- |
| 534 |
| 1546 |
| 2024 |
| 73104 |

Note. With a longer consideration, the result can also be derived solely from the requirement that the number $A B C$ and its square together contain each of the digits $0,1,2, \ldots, 7$ exactly once, because in the octal system, only this number has this property.