1995 AMC 8 Problem 21
Attempt Problem 21 of the 1995 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 1995 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
21.
A plastic snap-together cube has a protruding snap on one side and receptacle holes on the other five sides. What is the smallest number of these cubes that can be snapped together so that only receptacle holes are showing?
Answer: B
Small hint:
Each cube has one snap that must be hidden inside another cube’s hole
Big hint:
Try arranging the cubes in a small ring so every snap plugs into a neighbor
Solution:
Every cube’s single snap must be plugged into another cube’s hole to be hidden. Two cubes can share only one face, so two cubes cannot hide both snaps. With three cubes, hiding all three snaps would require a three-cube loop in which every pair shares a face, but three unit cubes cannot be pairwise face-adjacent. Thus one, two, or three cubes cannot work.
Four cubes can be arranged in a square ring, each snap fitting into the neighbor’s hole, so only receptacle holes show. The smallest number is
Thus, the correct answer is B .
Problem 21 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
- 1994 AMC 8
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- 1997 AMC 8
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- 1999 AMC 8
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