AMC8 2025

AMC8 2025 · Q15

AMC8 2025 · Q15. It mainly tests Basic counting (rules of product/sum), Casework.

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 $m+M$?

Kei画了一个$6$×$6$网格。他将$13$个单元正方形涂成银色，其余涂成金色。然后他沿垂直方向对折网格，形成重叠的单元正方形对。让$m$和$M$分别是金对金对的最小和最大可能数。$m+M$的值是多少？

(A) \ 12 \ 12

(B) \ 14 \ 14

(D) \ 18 \ 18

(E) \ 20 \ 20

Answer

Correct choice: (C)

正确答案：（C）

Solution

We can see that the least number of gold-on-gold pairs will be obtained when the $13$ silver squares are placed on the two sides so that they don't overlap when folded over (because then it will minimize the number of gold-on-golds). We can see that if we split them up $6$ and $7$ on both sides, and then fold it, the number of gold-on-golds will be $18-13 = 5$. The maximum number of gold-on-golds will be achieved when the silver squares overlap when folded over, which will increase the number of gold-on-golds. If we align 6 silver squares with each other on each side, and put the last one somewhere else, we get the maximum is $18 - 7 = 11$. Therefore, the answer is $11+5=\boxed{\textbf{(C)}~16}$.

我们可以看到，最少金对金对是在两侧放置$13$个银色正方形，使折叠时不重叠（从而最小化金对金对）时得到。如果两侧分$6$和$7$，折叠后金对金对数为$18-13 = 5$。最大金对金对是在银色正方形折叠时重叠，从而增加金对金对时得到。如果两侧各对齐$6$个银色正方形，剩下一个放别处，最大为$18 - 7 = 11$。因此，答案是$11+5=\boxed{\textbf{(C)}~16}$。

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