The Konigsberg School has assigned grades $1$ through $7$ to pods $A$ through $G$, one grade per pod. Some of the pods are connected by walkways, as shown in the figure below. The school noticed that each pair of connected pods has been assigned grades differing by $2$ or more grade levels. (For example, grades $1$ and $2$ will not be in pods directly connected by a walkway.) What is the sum of the grade levels assigned to pods $C$, $E$, and $F$?

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.