Some unit cubes are assembled to form a larger cube. Some of the faces of the larger cube are then painted. The cube is taken apart and it is found that 217 of the unit cubes have paint on them. What is the total number of unit cubes?
729 unit cubes
If we make a large cube which is 9 unit cubes on a side then it is comprised of 9x9x9 = 729 unit cubes.
The interior is a 7x7x7 cube, therefore, the surface is made up of 729 - 7x7x7 = 729 - 343 = 386 unit cubes, and each face is 9x9=81 unit cubes, so the interior square of each face is 7x7=49 unit cubes.
But we have to come up with 217 cubes, which is 169 cubes fewer than the 386 surface cubes.
Now, if we don't paint THREE sides of the large cube, that all share the same corner
, then we will avoid painting the interiors of the sides (3 x 49 = 147 unit cubes), 3 common sides (3 x 7 = 21 unit cubes) and 1 common corner (1 unit cube). Therefore, we have avoided painting (147 + 21 + 1 =) 169 unit cubes leaving 217 painted, which is what we were looking for!
Alternatively (and equivalently), if we paint THREE sides of the large cube, that all share the same corner
, then we paint 3x81 cubes, less the 3 2-side intersections -(3x8), less the 1 3-side intersection (counted twice) -2, = 243 - 24 - 2 = 217
painted, which is what we were looking for.
So far, we've proved that this is *a* solution....
To prove this is the *only* solution:
A 7-cube has 7x7x7 - 5x5x5 = 343 - 125 = 218 surface cubes, which is one too many, and we can't reduce the painted cubes by 1, so all solutions must be larger than a 7-cube.
If we go up to a 15x15 cube, one side is 225 cubes, so, we know all solutions, are smaller than a 15-cube
Also, we can rule out 14x14 (because we can't get to 217 with sides of 196 cubes).
We can rule out 13x13 (because we can't get to 217 with sides of 169).
We can rule out 12x12 (because we can't get to 217 with sides of 144).
Now we know the solution set lies (inclusively) between a 8-cube and an 11-cube.
With an 8 cube, the faces are 64 a piece... if we paint three faces with only two shared edges (must share at least two), we paint 64x3 - (8x2) = 192-16 = 176, too few. If we paint four faces with four shared edges (must share at least 4), then we get 64x4 - 4x8 = 256 - 32 = 216, too few, and if we paint a 5th face, we add another 36 painted cubes, fare too much. So we've eliminated that.
This leaves only 9, 10 and 11 cubes. We've already showed that a 9 cube will work.
10-cube has sides of 100 cubes. So, we must either paint two or three sides. Two sides will give us 200 or 190 painted cubes (too few), and three sides will give us 300 less either 20 or 28 (depending on how many shared edges), too many.
Similarly for an 11-cube, one side is only 121 painted cubes, and two sides gives us 242 or 231, too much.
This leaves us with a 9-cube being the only solution.