In the following square root, where each letter and dot represents a digit (equal letters represent the same digit, different letters represent different digits), write the appropriate digits in place of the letters and dots:

$$
\begin{aligned}
a b c d^{2}= & \\
& \ldots e \\
& \ldots c \\
& \ldots \cdots b \\
& \quad \ldots \ldots b
\end{aligned}
$$
I. Solution: When squaring the four-digit number, we obtained an eight-digit number. Since the square of 3000 is still only seven digits, therefore $a \geq 3$. On the other hand, in the squaring process, the square of the first digit is a single digit, so $a \leq 3$. Combining the two, $a=3$, and since $a^{2}=b$, $b=9$.

The second partial result was created by writing the second digit next to twice the first digit, and then multiplying the resulting number by the second digit: $69 \cdot 9=621$. Therefore, $e=1$.

According to the third partial result, the last digit of $c^{2}$ is $c$. Such a number can be $0, 1, 5$, and 6. If $c=0$, then this row would be all zeros, so we discard this. $c=1$ is ruled out because $e=1$. However, both $c=5$ and $c=6$ are valid.

In the fourth partial result, $d^{2}$ ends in 9. $d=3$ is not possible because $a=3$, so only $d=7$ is possible.

The letters and dots can be replaced by digits in two ways in the final result:

| $3957^{2}=$ | 9 | $3967^{2}=$ | 9 |
| ---: | :--- | ---: | :--- |
|  | 621 |  | 621 |
| 3925 |  | 4716 |  |
|  | $\frac{55349}{15657849}$ |  | 55489 |

Lajos Trón (Debrecen, Fazekas g. II. o. t.)

II. Solution: Since the sum of two single-digit numbers is at most 18, and according to the second partial result, $b^{2}$ ends in $e$, and there is no square number ending in 2, therefore $e$ can only be 1.

According to the second partial result, the last digit of $b^{2}$ is 1, and since $e=1$, $b$ cannot be 1, so $b=9$.

From $b=a^{2}$, $a=3$, and the solution proceeds in the same way as in the first solution.