$35 \cdot 7$ Divide each face (a total of six faces) of a $3 \cdot 3 \cdot 3$ cube into 9 smaller squares of the same size (a total of 54 smaller squares). Now, use red, yellow, and blue to color these smaller squares, such that adjacent squares sharing an edge cannot be the same color. Then, the maximum number of smaller squares that can be colored red is
(A) 28.
(B) 27.
(C) 22.
(D) 20.
("Zu Chongzhi Cup" Junior High School Mathematics Invitational Competition, 1989)
[Solution] We need to consider that small squares on the same face cannot be the same color if they share an edge, and also that small squares on adjacent faces cannot be the same color if they share an edge (ridge).

Since each small cube can only have one face painted red, this means there are a total of $3^{3}-1=26$ small cubes that can be painted red (excluding the central small cube). Therefore, options $(A)$ and $(B)$ can be ruled out.
There exists a method where 22 small cubes can be painted red (see the figure below):
The coloring method for the 1st and 3rd layers is the same, with a total of $9 \cdot 2=18$ small cubes that can be painted red; the 2nd layer can have 4 small cubes painted red, thus the total is $18+4=22$.
Therefore, the correct choice is $(C)$.