2025 AMC 8 第 15 题

先试着解答 2025 AMC 8 第 15 题,然后核对你的答案与视频与精心整理的解答,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 2025 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。

15.

凯画了一个 66 乘 66 的网格。他把 1313 个单位正方形涂成银色,其余涂成金色。然后凯沿竖直方向把网格对折,形成重叠的单位正方形对。设 mm 和 MM 分别为金色叠金色的对数的最小值和最大值。求 m+Mm+M。
Kei draws a 66-by-66 grid. He colors 1313 of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let mm and MM equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of m+M?m+M?
1212
1414
1616
1818
2020
答案:C
知识点:配对与分组极端原理
小提示:
折叠后形成 1818 对重叠方格
The fold makes 1818 overlapping pairs
大提示:
要最小化双金色对,先把金色方格尽量分散到不同对中
To minimize double-gold pairs, spread gold squares among pairs first
视频讲解:
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文字解答:
金色方格数为 6×6−13=36−13=23。 6 \times 6 - 13 = 36 - 13 = 23\text{。}折叠后,3636 个方格形成 1818 对重叠方格。
要让双金色对数最小,先把金色方格分散到所有对中。这用掉 1818 个金色方格,剩下 23−18=523 - 18 = 5 个必须与已有金色方格重叠,所以 m=5m = 5。
要让双金色对数最大,就尽量把 2323 个金色方格两两配对。最多可形成 M=11M = 11 对,并剩下 11 个金色方格,因为 23÷223 \div 2 商 1111 余 11。
答案是 m+M=5+11=16m + M = 5 + 11 = 16,选 C。
The number of gold squares is 6×6−13=36−13=23. 6 \times 6 - 13 = 36 - 13 = 23. The 3636 total squares overlap as 1818 pairs.
To minimize the number of pairs with two gold squares, the gold squares should first be spread out across all pairs. That uses up 1818 of them. The remaining 23−18=523 - 18 = 5 gold squares double-up and create a total of m=5m = 5 gold-on-gold pairs.
To maximize the number of pairs with two gold squares, the 2323 gold squares should first be paired up as much as possible. That can be done to create M=11M = 11 pairs, with 11 gold square left over, because 23÷223 \div 2 is 1111 with a remainder of 1.1.
The answer is m+M=5+11=16,m + M = 5 + 11 = 16, which is choice C.
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