2025 AMC 8 Problem 15

Attempt Problem 15 of the 2025 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

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
Answer: C
Concepts:pairing and groupingextremal argument
Small hint:
The fold makes 1818 overlapping pairs
Big hint:
To minimize double-gold pairs, spread gold squares among pairs first
Video solution:
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Written solution:
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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