Day 4 · Painted Cube¶
Edited by Charles
Combinatorics · Spatial reasoning
Original question on OpenQuant
Problem¶
A solid \(10\times10\times10\) cube is assembled from unit cubes. Paint coats its entire outer surface without entering gaps between the small cubes. How many unit cubes remain completely unpainted?
Core idea
A small cube stays unpainted exactly when it is strictly inside the large cube. Remove the outermost layer from both ends of every axis.
Short solution¶
1. Identify allowed positions along one axis. Number positions from \(1\) to \(10\). A cube at position \(1\) or \(10\) touches the exterior in that direction. Only positions \(2,3,\ldots,9\) can belong to the unpainted interior: eight choices.
2. Apply the restriction on all three axes. Every unpainted cube must have an interior position in the horizontal, vertical, and depth directions. The interior is an \(8\times8\times8\) cube.
Answer
Watch out
Subtracting \(6\cdot10^2\) from \(10^3\) double-counts edge cubes and triple-counts corner cubes. Counting the interior directly avoids overlapping faces.
Check by classifying the painted cubes
There are eight corners with three painted faces, \(12(10-2)=96\) edge cubes with two painted faces, and \(6(10-2)^2=384\) face-interior cubes with one painted face.
More generally, an \(n\times n\times n\) solid cube has \(\max(n-2,0)^3\) completely unpainted unit cubes.