2022 AMC 8 第 20 题

先试着解答 2022 AMC 8 第 20 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2022 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

下方网格要填入整数,使得每一行的和与每一列的和都相同。四个数字缺失。左下角的数字 xx 大于另外三个缺失数字。xx 的最小可能值是多少?
The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number xx in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of x?x?
−1-1
55
66
88
99
答案:D
知识点:幻方不等式
小提示:
已完整的行给出每行和每列的共同和
The completed row and column sums must match the top row
大提示:
用 xx 表示另外三个缺失项
Express the other missing entries in terms of xx
解答:
由顶行可知每行、每列的和都必须为 1212。
在第一列中,xx 上方的缺失数为 14−x14-x。在底行中,xx 右边的缺失数为 4−x4-x。中间行剩下的缺失数为 x−1x-1。
因为 xx 大于另外三个缺失数,需要 x>14−xx>14-x、x>4−xx>4-x、x>x−1x> x-1。最强条件是 x>7x>7,所以最小整数为 88。
正确答案是 D。
Adding the numbers in the top row shows that every row and column must have sum 1212.
In the first column, the missing number above xx is 14−x14-x. In the bottom row, the missing number to the right of xx is 4−x4-x. In the middle row, the remaining missing number is x−1x-1.
Since xx is larger than the other missing numbers, we need x>14−xx>14-x, x>4−xx>4-x, and x>x−1x> x-1. The strongest condition is x>7x>7, so the smallest possible integer value is 88.
Thus, the correct answer is D.
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