AMC8 2025

AMC8 2025 · Q21

AMC8 2025 · Q21. It mainly tests Geometry misc, Logic puzzles.

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 1 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 $F$?

柯尼斯堡学校将1到7的年级分配给荚A到G，每个荚一个年级。一些荚通过人行道连接，如下图所示。学校注意到，每对相连的荚分配的年级相差1个或更多年级。（例如，1和2年级不会在直接连接的人行道荚中。）C、E和F荚分配的年级水平之和是多少？

(A) 12 12

(B) 13 13

(D) 15 15

(E) 16 16

Answer

Correct choice: (A)

正确答案：（A）

Solution

The key observation for this solution is to observe that pods $C$ and $F$ both have degree 5; that is, they are connected to five other pods. This implies that $C$ and $F$ must contain grades 1 and 7 in either order (if $C$ or $F$ contained any of grades 2, 3, 4, 5, or 6, then there would only be four possible grades for five pods, a contradiction). It does not matter which grade $C$ and $F$ gets (see Solution 2 below), so we will assume that pod $C$ is assigned grade 1, and pod $F$ is assigned grade 7. Next, pod $D$ is the only pod which is not adjacent to pod $F$, so pod $D$ must be assigned grade 6. Similarly, pod $G$ must be assigned grade 2. Lastly, we need to assign grades 3, 4, and 5 to pods $A$, $B$, and $E$. Note that $A$ and $B$ are adjacent; therefore, pods $A$ and $B$ must contain grades 3 and 5. We conclude that $E$ must be assigned grade 4. The requested sum is $1+4+7 = \mathbf{(A)\,} 12$.

本题解法的关键观察是荚C和F的度均为5，即它们连接到其他五个荚。这意味着C和F必须包含1和7（顺序任意）（如果C或F包含2、3、4、5或6中的任何一个，那么对于五个荚只有四个可能的年级，矛盾）。C和F哪个得到哪个年级无关紧要（见下面的解法2），因此我们假设荚C分配1年级，荚F分配7年级。接下来，荚D是唯一不与荚F相邻的荚，因此荚D必须分配6年级。类似地，荚G必须分配2年级。最后，我们需要将3、4和5年级分配给荚A、B和E。注意A和B相邻；因此荚A和B必须包含3和5。我们得出E必须分配4年级。要求的和是$1+4+7 = \mathbf{(A)\,} 12$。

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