Cutting a Cube into Smaller Ones

Problem

Cutting a Cube into Smaller Onese, problem

Solutions, 1

For $15:$ see it as a $2\times 2\times 2$ cube. Cut one of the small cube into $8$ in the same manner. The result: $8-1+8=15.$

For $20:$ see it as a $3\times 3\times 3$ cube. Carve out a corner $2\times 2\times 2$ cube. Cut the rest into $19=27-8$ $1\times 1\times 1$ cubes. The result: $27-8+1$ cubes.

For $22:$ see it as a $2\times 2\times 2$ cube. Cut two of the small cube into $8$ in the same manner. The result: $8-2+2\cdot 8=22.$

For $45:$ see it as a $4\times 4\times 4$ cube. Carve out a corner $3\times 3\times 3$ cube. Cut one of the small $1\times 1\times 1$ cubes into $8.$ The result: $64-27+1-1+8=45$ cubes.

Solutions, 2

For $15:$ see it as a $4\times 4\times 4$ cube. Cut it into eight $2\times 2\times 2$ and cut one of the latter into $2\times 2\times 2$ smaller ones. The result: $4^3-7\cdot 2^3+7=15.$

For $22:$ see it as a $5\times 5\times 5$ cube. Carve a $3\times 3\times 3$ cube, eleven $2\times 2\times 2$ cubes and ten $1\times 1\times 1$ cubes.

For $45:$ see it as a $6\times 6\times 6$ cube. Carve out two adjacent corner $3\times 3\times 3$ cubes. Cut out eighteen $2\times 2\times 2$ cubes and one of them and the remaining space into twenty six $1\times 1\times 1$ cubes. The result: $2+17+26=45$ cubes of the total volume $2\cdot 27+17\cdot 8+26=216.$

Illustrated Solutions, 3

Cutting a Cube into Smaller Ones, solution 3

Acknowledgment

R. Honsberger's Mathematical Chestnuts from Around the World (MAA, 2001) features two discussions on the question of cutting a cube into smaller ones. The first (pp 92-94) handles the question of always having at least two cubes of the same size; the second one (p 191) deals with the question from the 1987 Hungarian Mathematical Olympiad for 11-year-olds where the task was to cut a cube into $20$ smaller ones.

Second solutions are by Viral Doshi; The illustrated solution is by Adrián Begoña.

 

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