# 1985 AHSME Problems/Problem 20

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## Problem

A wooden cube with edge length $n$ units (where $n$ is an integer $>2$) is painted black all over. By slices parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit length. If the number of smaller cubes with just one face painted black is equal to the number of smaller cubes completely free of paint, what is $n$?

$\mathrm{(A)\ } 5 \qquad \mathrm{(B) \ }6 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 8 \qquad \mathrm{(E) \ }\text{none of these}$

## Solution

Notice that if we remove the outer layer of unit cubes from the entire cube, we're left with a smaller cube of side length $n-2$. Notice also that this contains all of the unpainted cubes and nothing else, so there are $(n-2)^3$ unpainted cubes. Also notice that if we take one face of the cube and remove the outer edge, we're left with a square of side length $n-2$ containing all of the cubes on that face with exactly one face painted. This can be done for the other $5$ faces as well, for a total of $6(n-2)^2$ cubes with only one face painted.

$6(n-2)^2=(n-2)^3$

$6=n-2$

$n=8, \boxed{\text{D}}$.