Below is a square root calculation scheme written out in full detail, where the letter $x$ represents missing digits. What is the number under the root?

$$
\begin{aligned}
& \sqrt{x x x 8 x x}=x x x \\
& \underline{x x} \\
& x x x: x x \cdot x \\
& \quad \underline{x x} \\
& \quad x x x x: x x x \cdot x \\
& \quad \underline{x x x x} \\
& 1
\end{aligned}
$$
```
xxA}=xx
xx
xxx:xx.x
                            \(\underline{x}\)
\(x x x x: x x x \cdot x\)
\(\underline{x x x x}\)
1
```

The first digit of the square root is at least 4, because according to the second row of the schema, its square is a two-digit number. On the other hand, this digit is at most 4, because according to the third row, twice the preceding digit is a single-digit number. Thus, the first digit of the root is 4, and the complete divisor in the third row is a number between 80 and 89.

The second digit of the root is 1, because if it were more, then the subtrahend in the fourth row would be a three-digit number, but it cannot be less, i.e., 0, because then the remainder in the fifth row would have to match the dividend in the third row. Thus, the subtrahend in the fourth row is 81, the remainder in the fifth row is \(\overline{x 7}\), the dividend is \(\overline{x 7 x x}\), and the temporary divisor is 82.

The third digit of the root is at least 2, because the subtrahend in the sixth row is more digits than \(\overline{82 x}\), the complete divisor, and at most 9. The hundreds digit of the subtrahend in the sixth row is either 7, like the hundreds digit of the dividend in the fifth row, or 6, but by adding 1 to the subtrahend, it becomes 7, which can only happen if the subtrahend is in the form \(\overline{x 699}\). The subtrahend is the product \(\overline{82 x} \cdot x\) (where \(x\) stands for the same digit in both places). Here, \(\overline{2 x} \cdot x < 300\). Thus, the hundreds digit of the product can be 6 only if \(8 \cdot x\) (which is even) ends in 6 with no remainder, or ends in 4 with a remainder of 2. The hundreds digit can be 7 only if \(8 \cdot x\) ends in 8 with a remainder of 1. 6 is the double of 8 at 2 and 7 times, and 4 is the triple and 8 times of 3. Thus, the subtrahend is \(822 \cdot 2 = 1644\), \(823 \cdot 3 = 2469\), \(827 \cdot 7 = 5789\), or \(828 \cdot 8 = 6624\). None of these are in the form \(\overline{x 699}\), and the hundreds digit is only 7 in the third one, so the square root could only have been taken from \(417^2 + 1\). In this case, the square root extraction indeed proceeds according to the schema provided.

Ferenczy György (Budapest, I. István g. I. o. t.)
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