Problem 24 · 1991 AJHSME
Stretch
Geometry & Measurement
volume-decomposition
A cube of edge 3 cm is cut into N smaller cubes, not all the same size. If the edge of each smaller cube is a whole number of centimeters, then N =
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Answer: E — 20.
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Hint 1 of 3
What sizes of smaller cube can even exist here? Edges must be whole centimeters and SMALLER than 3, so only edge 1 or edge 2 are allowed. The phrase "not all the same size" means you can't just use twenty-seven 1-cubes — you must include at least one 2-cube.
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Hint 2 of 3
A 2×2×2 cube needs 2 cm of room in every direction, and the big cube is only 3 cm wide — so how many 2-cubes can you possibly fit side by side? Place that many, then fill every remaining gap with 1-cubes.
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Hint 3 of 3
Total volume is 3³ = 27. Subtract the volume of the 2-cube(s) you placed; whatever's left is made of unit cubes (volume 1 each). Then add up the piece COUNT.
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Approach: only 1- and 2-cubes are allowed; place the few big ones, count the rest
- The big cube has volume 3 × 3 × 3 = 27. Smaller whole-cm cubes can only have edge 1 or 2 (edge 3 would be the whole cube). And "not all the same size" forces at least one 2-cube into the mix.
- How many 2-cubes fit? Each needs 2 cm along every edge, but the box is only 3 cm wide, so two of them can't sit side by side in any direction — only ONE 2-cube fits.
- That 2-cube uses 8 of the 27 cubic cm, leaving 27 − 8 = 19 cubic cm, all filled by unit cubes — that's 19 of them.
- Total pieces: 1 + 19 = 20, and they're not all the same size. ✓
- Why this transfers: in dissection problems, first list which piece sizes are even possible, then use the container's dimensions to cap how many of the big pieces fit — the leftover volume forces the count of small pieces. Counting volume and counting pieces are different questions; here we needed the piece count.
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