1994 AMC 8 Problem 24
Attempt Problem 24 of the 1994 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1994 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
24.
A by square is divided into four by squares. Each of the small squares is to be painted either green or red. In how many different ways can the painting be accomplished so that no green square shares its top or right side with any red square? There may be as few as zero or as many as four small green squares.
Answer: B
Small hint:
If a square is green, then the square directly above it and the square directly to its right must also be green
Big hint:
This forces the green squares to cluster toward the top-right; count by how many squares are green
Solution:
The rule says a green square cannot have a red square on its top or right side, so any green square forces the squares above and to its right to be green as well. The green squares must therefore cluster toward the top-right corner.
The valid colorings are: all four red; only the top-right green; the whole top row green; the whole right column green; all green except the bottom-left; and all four green. That is colorings.
Thus, the correct answer is B .
Problem 24 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
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