【Question 14】
As shown in the figure, fill in the table according to the following requirements:
(1) Enter a number in each small square, and this number must be a natural number less than 10;
(2) Small squares within the same region must be filled with the same number (dashed lines indicate the interior of the same region);
(3) The small squares in the leftmost column cannot be filled with the number 0;
(4) The four digits in the first row from top to bottom, read from left to right, form a four-digit number, which can be expressed as $p^{q}$, where $p$ and $q$ are both prime numbers;
(5) The four digits in the second row from top to bottom, read from left to right, form a four-digit number, which can be expressed as the product of two consecutive prime numbers (for example, 3 and 5 are consecutive prime numbers);
(6) The four digits in the third row from top to bottom, read from left to right, form a four-digit number, which is a perfect square;
(7) The four digits in the fourth row from top to bottom, read from left to right, form a four-digit number, which is a multiple of 37.
【Analysis and Solution】
Comprehensive number theory.
Each small square is filled with a letter, and squares within the same area are filled with letters, where the same letter represents the same number, and different letters can represent either different or the same numbers. From (6), we get that $\overline{E E G G}$ is a perfect square;

The last two digits of a perfect square can be $00, 25$, odd 6, even 1, $\overline{\text {even } 4}$, even 9; if the last two digits of a perfect square are the same, then they can only be 00 or 44; and $\overline{E E G G}=\overline{E 0 G} \times 11$ is a multiple of 11;

Thus, $\overline{E E G G}$ is a multiple of $11^{2}$;
$\overline{E 0 G}=\overline{E E G G} \div 11$ is a multiple of 11;

By the divisibility rule of 11, we get $(E+G)-0=E+G$ is a multiple of 11;
$E+G=11$;

Therefore, $G$ cannot be 0, it can only be $G=4, E=11-G=11-4=7$; and $7744=88^{2}$.
From (7), we get that $\overline{E H G G}=\overline{7 H 44}$ is a multiple of 37;
$$
999=3^{3} \times 37 \text {; }
$$

Since $\overline{7 H 44}$ is a multiple of 37;

Then $\overline{7 H 440}$ is also a multiple of 37;

By the divisibility rule of 37, $\overline{7 H}+440=510+H$ is a multiple of 37;
$$
\begin{array}{l}
510 \div 37=13 \cdots \cdots 29 ; \\
H=37-29=8 .
\end{array}
$$

From (5), we get that $\overline{E F C D}=\overline{7 F C D}$ can be expressed as the product of two consecutive prime numbers;
We estimate using multiples of ten: $80^{2}=6400, 90^{2}=8100$;
Further calculations yield, $79 \times 83=6557, 83 \times 89=7387, 89 \times 97=8633$;
Therefore, $\overline{E F C D}=\overline{7 F C D}$ can only be $83 \times 89=7387$;
$F=3, \quad C=8, \quad D=7$.
From (4), we get that $\overline{A B C D}=\overline{A B 87}=p^{q}$, where $p$ and $q$ are both prime numbers;
First, the last digit of $p$ can only be $3, 7$;
$q$ cannot be too large;

Notice that $3^{11}>9999$, so $q=2, 3, 5, 7$.
When $q=2$, because the last digit of a perfect square can be $0, 1, 4, 9, 6, 5$, it cannot be $2, 3, 7, 8$; hence, it does not exist.

When $q=3$, the last digit of $p$ can only be 3, after trying, $3^{3}=27, 13^{3}=2197, 23^{3}>9999$, it does not exist;

When $q=5$, the last digit of $p$ can only be 7, after trying, $7^{5}>9999$, it does not exist;

When $q=7$, the last digit of $p$ can only be 3, after trying, $3^{7}=2187, 13^{7}>9999; 3^{7}=2187$ fits;

Therefore, $\overline{A B C D}=\overline{A B 87}=2187$;
$$
A=2, \quad B=1 \text { . }
$$