$28 \cdot 1$ In a column of squares on the right,
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline$A$ & 9 & $H$ & $M$ & $O$ & $X$ & 7 \\
\hline
\end{tabular}
each square, except for 9 and 7, represents a number. It is known that the sum of any 3 consecutive squares is 19. Then $A+H+M+O$ equals
(A) 21 .
(B) 23 .
(C) 25 .
(D) 26 .
(Anhui Province Junior High School Mathematics Competition, 1997)
［Solution] From the given, we easily know that $A+H=10$.
$$
\because \quad M+O+X=O+X+7, \quad \therefore \quad M=7 \text { : }
$$

Similarly,
$$
\begin{array}{ll}
\text { Similarly, } O=9 . \\
\therefore & A+H+M+O=26 .
\end{array}
$$

Therefore, the answer is (D).