2025 AMC 8 Problems/Problem 15

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Problem

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 equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of $m+M$?

$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 20$

Solution 1

First, we note that there are $18$ "pairs" of squares folded on top of each other after the folding. The minimum number of gold pairs occurs when silver squares occupy the maximum number of pairs, and the maximum number of gold pairs occurs when silver squares occupy the minimum number of pairs. The former case occurs when all $13$ silver squares are placed in different pairs, resulting in $18-13=5$ gold pairs. The latter case occurs when the silver squares are paired up as much as possible, resulting in $6$ complete pairs and another square occupying another pair slot. Then there are $18-7=11$ gold pairs. Our answer is $11+5=\boxed{\textbf{(C)}~16}$. ~cxsmi

Solution 2

We first observe that there are $6^2 - 13 = 23$ gold squares. To find $M$, we can note that $\frac{23}{2} = 11 \frac{1}{2}$, so we can have at most $11$ pairs (which can be constructed). To find $m$, we fill one side up entirely with gold squares, leaving $5$ left for the other side. From this position, no matter where we move a square, there will always be $5$ overlapping gold-on-gold pairs, so $m = 5$. It follows that $m + M = 5 + 11 = \boxed{\textbf{(C)\ 16}}.$ I would like to state that there is a video below if you do not understand the solution.

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI You can contact him through his Youtube Channel

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

See Also

2025 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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