2025 AMC 8 第 21 题
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所有题目均经美国数学协会（MAA）官方合法授权使用。
21.
哥尼斯堡学校把 到 年级分别分配给 到 七个教学舱，每个教学舱一个年级。部分教学舱由步道连接，如下图所示。学校注意到每一对相连教学舱的年级差都至少为 个年级。（例如， 年级和 年级不会在由步道直接相连的教学舱中。）分配给教学舱 、、 的年级之和是多少？
The Konigsberg School has assigned grades through to pods through 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 or more grade levels. (For example, grades and will not be in pods directly connected by a walkway.) What is the sum of the grade levels assigned to pods and
小提示：
四个两两相连的教学舱必须得到年级 、、、
The four mutually connected pods must get grades and
大提示：
利用教学舱 限制较小和较大的年级可以放在哪里
Use pod to force where small and large grades can go
视频讲解：
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文字解答：
教学舱 、、、 两两相连。 到 中四个两两至少相差 的数只能是 。因此 、、 使用偶数年级 。
教学舱 与 和 相连。若 ，则 和 必须是 和 ，留下 和 为 和 。但 与 和 相连，剩下的偶数年级都会与其中一个只差 ，所以 不能是 。
若 ，则 和 必须是 和 。若 ，则 不能是 或 ，所以 。并且 不能是 ，否则 同样不能是 或 。因此 ，，得到 。
的情况对称，得到 、、。和同样是 。
所以正确答案是 A。
Pods and are all pairwise connected. Four numbers from through that are all at least apart must be . Thus and get the even grades .
Pod is connected to and . If , then and would have to be and , leaving and as and . But then , which is connected to and , cannot be either remaining even grade without being only away from one of them. So cannot be .
If , then and must be and . If , then cannot be or , so . Also, cannot be , since then again cannot be or . Therefore and , giving .
The case is symmetric, giving and . The same sum is .
Thus, A is the correct answer.
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