1991 AJHSME Problems/Problem 24

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Problem

A cube of edge $3$ cm is cut into $N$ smaller cubes, not all the same size. If the edge of each of the smaller cubes is a whole number of centimeters, then $N=$

$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$

Solution

If none of the cubes have edge length $2$ , then all of the cubes have edge length $1$ , meaning they all are the same size, a contradiction.

It is clearly impossible to split a cube of edge $3$ into two or more cubes of edge $2$ with extra unit cubes, so there is one $2\times 2\times 2$ cube and $3^3-2^3=19$ unit cubes.

The total number of cubes, $N$ , is $1+19=20\rightarrow \boxed{\text{E}}$ .

1991 AJHSME (Problems • Resources)
Preceded by Problem 23	Followed by Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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1991 AJHSME Problems/Problem 24

From AoPSWiki

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