Problem 15 · 2003 AMC 8
Hard
Geometry & Measurement
spatial-reasoning
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Answer: B — 4 cubes.
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Hint 1 of 2
For a MINIMUM, make each cube do double duty — one cube can show up in both the front view and the side view at once.
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Hint 2 of 2
Build the L-shaped front view with 3 cubes first, then add only what the side view still demands.
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Approach: reuse cubes across both views; add only what's missing
- Minimizing means letting cubes count toward both pictures. Start with the front view, an L of 3 squares — that needs at least 3 cubes, sitting in one flat plane.
- Now check the side view: it's also an L with depth, meaning something must sit behind the front row. Those 3 cubes alone give a side view only 1 cube deep, which is wrong. Adding one cube behind the corner fixes the side view — and that 4th cube touches the others, so the "every cube shares a face" rule holds.
- No fewer than 4 can cover both an L front and an L side, so the minimum is 4 cubes.
- You'll see this again: in "fewest cubes for these views" problems, the answer is driven by where the two views disagree — build the bigger view, then patch only the parts the other view still forces.
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