Problem 7 · 1987 AJHSME
Medium
Geometry & Measurement
count-by-sub-cube-class
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Answer: C — 20.
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Hint 1 of 3
Counting the shaded ones directly is messy. Instead ask the reverse: which small cubes manage to stay completely unshaded?
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Hint 2 of 3
Every one of the 27 small cubes is one of just four types by position: 8 corners, 12 edges, 6 face-centers, 1 hidden inside. Sort by type instead of counting square by square.
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Hint 3 of 3
The checkerboard leaves only each face's center square white. The single cube directly behind that center square is the only one on that face that can dodge all shading.
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Approach: count by sub-cube class (and count the easy ones)
- Sort the 27 small cubes by position: 8 sit at corners (3 faces showing), 12 sit on edges (2 faces showing), 6 sit at the center of a face (1 face showing), and 1 is buried inside (0 faces).
- It's far easier to count the cubes with NO shading. The pattern leaves only the center square of each face white, so the only candidates to escape are the 6 face-center cubes plus the 1 internal cube = 7 unshaded.
- Everything else is shaded: 27 − 7 = 20.
- Why this transfers: "count the complement" — when the thing you want is scattered and the leftover is tidy, count the leftover and subtract from the whole.
Another way — add the shaded classes directly:
- All 8 corner cubes and all 12 edge cubes show at least one shaded square (they touch a non-center square of some face).
- 8 + 12 = 20, confirming the complement count.
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