哥尼斯堡学校将 $1$ 至 $7$ 年级的学生分配到 $A$ 至 $G$ 教学区，每个教学区分配一个年级。如下图所示，部分教学区通过走廊相连。学校注意到，每两个相连的教学区分配的年级数之差不低于 $2$（例如，$1$ 年级和 $2$ 年级的教学区不会直接通过走廊相连）。请问，分配到 $C$、$E$ 和 $F$ 教学区的年级数总和是多少？

A. 12

B. 13

C. 14

D. 15

E. 16

1. 2025 AMC 8 官方解析

Answer (A): Pods $C$ and $F$ have five adjoining pods each, so they must be assigned grades $1$ and $7$ in some order because all other grades can have at most four adjoining pods. Suppose grade $1$ is assigned to pod $C$ and grade $7$ to pod $F$. Then the only possible location for grade $2$ is pod $G$ because all other pods are connected to grade $1$. Similarly grade $6$ must be in pod $D$. That leaves grades $3$, $4$, and $5$ for pods $A$, $B$, and $E$. Because pods $A$ and $B$ are connected, pod $E$ must contain grade $4$. Then grades $5$ and $3$ can be assigned to pods $A$ and $B$, respectively, as shown below.

By a similar argument, if grade $7$ is assigned to pod $C$ and grade $1$ is assigned to pod $F$, the solution will be similar, as shown below. Again grade $4$ will be assigned to pod $E$.

For either solution, the sum of the grade levels in pods $C$, $E$, and $F$ is $1 + 4 + 7 = 12$. OR

For every possible assignment of grade levels to pods, there is an alternative assignment given by reversing the order of the grades $1$ through $7$. This replaces $C$, $E$, and $F$ by $8 - C$, $8 - E$, and $8 - F$, respectively. Assuming that there is exactly one possible value for $C + E + F$, both assignments give the same sum. This means $C + E + F = (8 - C) + (8 - E) + (8 - F) = 24 - (C + E + F)$, so $2 \cdot (C + E + F) = 24$, and $C + E + F = 12$.

Note: This problem is based on the Digit-Placing Problem (1.8) in Martin Gardner's The Colossal Book of Short Puzzles and Problems, W.W. Norton, 2006.