Question

Kei 画了一个 $6$ 乘 $6$ 的网格。他把 $13$ 个单位正方形涂成银色，其余涂成金色。然后 Kei 沿竖直方向把网格对折，形成重叠单位正方形对。设 $m$ 和 $M$ 分别为金色叠金色的对数的最小值和最大值。求 $m+M$ 。

Kei draws a $6$ -by- $6$ grid. He colors $13$ 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 $m$ and $M$ equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M?$

$12$

$14$

$16$

$18$

$20$

答案：C

知识点：配对与分组极端原理

难度评级：1480

视频讲解：

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文字解答：

金色方格数为折叠后， $36$ 个方格形成 $18$ 对重叠方格。 $6 \times 6 - 13 = 36 - 13 = 23.$

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 $18$ 个金色方格，剩下 $23 - 18 = 5$ 个必须与已有金色方格重叠，所以 $m = 5$ 。

要让双金色对数最大，就尽量把 $23$ 个金色方格两两配对。最多可形成 $M = 11$ 对，并剩下 $1$ 个金色方格，因为 $23 \div 2$ 商 $11$ 余 $1$ 。

答案是 $m + M = 5 + 11 = 16$ ，选 C。

The number of gold squares is $6 \times 6 - 13 = 36 - 13 = 23.$ The $36$ total squares overlap as $18$ 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 $18$ of them. The remaining $23 - 18 = 5$ gold squares double-up and create a total of $m = 5$ gold-on-gold pairs.

To maximize the number of pairs with two gold squares, the $23$ gold squares should first be paired up as much as possible. That can be done to create $M = 11$ pairs, with $1$ gold square left over, because $23 \div 2$ is $11$ with a remainder of $1.$

The answer is $m + M = 5 + 11 = 16,$ which is choice C.

2025 AMC 8 第 15 题

答案：C

难度评级：1480

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