2026 AMC 8 Problem 15
Attempt Problem 15 of the 2026 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 2026 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
15.
Elijah has a large collection of identical wooden cubes which are plain on faces and shaded on faces that share an edge. He glues some cubes together face-to-face. The figure below shows cubes being glued together, leaving shaded faces visible. What is the fewest number of cubes that he could glue together to ensure that no shaded faces are visible, no matter how he rotates the figure?
Answer: A
Small hint:
Every shaded face must be glued to another cube
Big hint:
With only cubes, some cube has at most one face-neighbor
Video solution:
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Written solution:
For no shaded face to be visible, each cube must have both of its shaded faces glued to neighboring cubes, so each cube needs two face-neighbors. With only three cubes, the face-adjacency graph is a path, so an end cube has only one face-neighbor. Four cubes arranged as a square can each be oriented with its two shaded faces pointing inward, so cubes suffice.
Problem 15 in other years
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