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# Difference between revisions of "1998 AHSME Problems/Problem 27"

## Problem

A $9 \times 9 \times 9$ cube is composed of twenty-seven $3 \times 3 \times 3$ cubes. The big cube is ‘tunneled’ as follows: First, the six $3 \times 3 \times 3$ cubes which make up the center of each face as well as the center $3 \times 3 \times 3$ cube are removed. Second, each of the twenty remaining $3 \times 3 \times 3$ cubes is diminished in the same way. That is, the center facial unit cubes as well as each center cube are removed. The surface area of the final figure is:

$\mathrm{(A)}\ 384 \qquad\mathrm{(B)}\ 729 \qquad\mathrm{(C)}\ 864 \qquad\mathrm{(D)}\ 1024 \qquad\mathrm{(E)}\ 1056$

## Solution

### Solution 1

Each $3 \times 3 \times 3$ cube has eight faces on each side, for a surface area of $6 \cdot 8 \cdot (1 \cdot 1) = 48$ on the outside. Each face also has to count the surface area in the inside area of the removed cube, for an additional surface area of $6 \cdot 4 \cdot (1 \cdot 1) = 24$. Thus the total surface area for each $3 \times 3 \times 3$ is $72$.

There are $20$ of these cubes, for an area of $72 \cdot 20 = 1440$. However, many of the cubes share a common face; each corner $3\times 3\times 3$ cube has three hidden faces and each edge cube has two hidden faces, for a total of $8\cdot 3 + 12\cdot 2 = 48$ hidden faces. Each hidden face has a surface area of $8$, so the surface area of the final figure is $1440 - 48 \cdot 8 = 1056 \Rightarrow \mathrm{(E)}$.

### Solution 2

After the first step, twenty $3 \times 3 \times 3$ cubes remain, with $8$ corner cubes and $12$ edge cubes. Each one of these $3 \times 3 \times 3$ corner cubes contributes $27$ square units of area, and each edge cube contributes $36$ square units of area.

The second stage takes away $3$ square units of area ($1$ for each exposed face) from each of the eight $3 \times 3 \times 3$ corner cubes, and adds an additional $24$ more units from the center facial cubes removed. Similarly, the twelve $3\times 3\times 3$ edge cubes each lose $4$ square units but gain $24$ units. Thus, the total surface area is $$8 \cdot (27 - 3 + 24) + 12 \cdot (36 - 4 + 24) = 1056$$

 1998 AHSME (Problems • Answer Key • Resources) Preceded byProblem 26 Followed byProblem 28 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 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions