Initial 275 Given $\qquad$
$$
\begin{array}{l}
23 \times \overline{a b c}=4 \times \overline{p q r}, 16 \times \overline{r p q}=5 \times \overline{x y z}, \\
22 \times \overline{z x y}=7 \times c a b,
\end{array}
$$

where $\overline{a b c} 、 \overline{p r} 、 \overline{x y z}$ represent three-digit numbers, and the same letters represent the same digits. Solve this number puzzle.
Let the three equations in the problem be denoted as (1), (2), and (3).
From equation (1), it is easy to see that $a=1$;
From equation (2), it is easy to see that $1 \leqslant r \leqslant 3$;
From equation (3), it is easy to see that $1 \leqslant z \leqslant 3$.
Furthermore, from equation (1), we know $2 \mid c$, from equation (2), we know $2 \mid z$, and from equation (3), we know $2 \mid b$.
Therefore, $z=2$.
From equation (3), $11 \mid \overline{c a b}$, i.e., $11 \mid \overline{c 1 b}$, so
$b+c=12$.
From equation (3), we get $22 \times \overline{2 x y}=7 \times \overline{c 1 b}$, so $(b, c)=(4,8)$ or $(6,6)$.
From equation (1), it is easy to see that $4 \mid \overline{b c}$, so $\overline{b c}=48, \overline{a b c}=148$.
From equation (1), we get $\overline{p q r}=23 \times 148 \div 4=851$.
From equation (2), we get $\overline{x y z}=16 \times 185 \div 5=592$.
Thus, the unique solution to this number puzzle is
$$
\left\{\begin{array}{l}
23 \times 148=4 \times 851, \\
16 \times 185=5 \times 592, \\
22 \times 259=7 \times 814 .
\end{array}\right.
$$
(Wu Kang, School of Mathematical Sciences, South China Normal University, 510631 Zhou Yi, Graduate Student, School of Mathematical Sciences, South China Normal University, 08 Level, 510631)