1991 AMC 8 Problem 24

Attempt Problem 24 of the 1991 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 1991 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

24.

A cube of edge 33 cm is cut into NN smaller cubes, not all the same size. If the edge of each of the smaller cubes is a whole number of centimeters, then N=N =
44
88
1212
1616
2020
Answer: E
Concepts:cube geometryvolume
Small hint:
The smaller cubes must have whole-number edges, so each is edge 11 or edge 22
Big hint:
Only one edge-22 cube fits; the rest are unit cubes filling the remaining volume 27−827 - 8
Solution:
The 3×3×33 \times 3 \times 3 cube has volume 27.27. Since the cubes are smaller and have whole-number edge lengths, their edges are 11 or 22. Two edge-22 cubes cannot be disjoint inside the original cube: separating them would require a total span of at least 44 in some direction. Thus at most one edge-22 cube fits.
The remaining 27−8=1927 - 8 = 19 of volume is filled by 1919 unit cubes. That is 1+19=201 + 19 = 20 cubes.
Thus, the correct answer is E .
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