You have a bag with many 3x3 Rubik's cubes . After solving all but one of them, you decide to play a new game.

You take the unsolved cube, and start placing solved ones around it.

The question: What is the maximum number of solved cubes that can touch the unsolved cube along its sides(fully or partially) ?

enter image description here


  • You can not remove any edge or corner piece!
  • All cubes are of same dimensions.
  • The solved cubes should touch the face of unsolved cube, so touching edges or at corner points do not count.
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    $\begingroup$ Does it have to be physically possible to put them in the proper position given the constraints of gravity and only having two hands? Or can we assume that I have the power to position cubes in space and have them remain in that position? $\endgroup$ – VictorHenry Jun 28 '16 at 20:06
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    $\begingroup$ You mention you have to touch the face of the unsolved cube. Does that also assume just using the faces of the solved cubes around it? $\endgroup$ – gtwebb Jun 28 '16 at 20:12
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    $\begingroup$ I don't know if this helps, but the question doesn't say the unsolved cube has to be untwisted. (If it is twisted, it wouldn't be a "cube" but it still would be a "rubik's cube", though even with a twist I currently don't see how we could eek another touch out of it.) $\endgroup$ – TTT Jun 28 '16 at 21:27
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    $\begingroup$ Why do we care that they are Rubik's cubes and one is unsolved? Isn't this the same question for solid cubes? You have not specified any restriction based on the color of the cubies. What you call sides are usually faces-you want an area of contact,not a point or line. -1 $\endgroup$ – Ross Millikan Jun 29 '16 at 3:55
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    $\begingroup$ Why is the puzzle using Rubik cubes and not plain cubes? I think I've seen this problem before (not in this site, in an old Scientific American, in Gardner's column). The answer was 24 or 26, if my memory is right. $\endgroup$ – ypercubeᵀᴹ Jun 29 '16 at 12:40

The answer is

at least $24$ (not sure if it's been proven no more can fit - it could be up to $26$ the bound for any touch)


We can fit $7$ on each of two opposite sides such that one side can still fit $6$, the side opposite that can fit $2$ and the remaining sides can have $1$ each (completely aligned) $7+7+6+2+1+1=24$

It's a tight squeeze ($6$ of the $7$s are just overlapping at the corners and most of the $6$ are on only a little more of the surface.

First one of the sides with $7$.
The black square is a side of the unsolved Rubik's cube.
Place a second cube (blue square) such that one corner, $p$ is not protruding an edge, but is as tight as possible to it, while protruding near both sides of the corner adjacent to, and on the other side of, the corner $p$ is nearest (top-right here) and shifted just slightly from producing a symmetric silhouette (up in the picture), to make a little extra room at the tiny available corner areas (top-right and bottom-right here).
Like this (but tighter to square, as we shall see):
first placement of 7
Now we can place the next $6$ of the $7$ in conjoined pairs such that their join overhangs three of the available corner spaces (the one closest to $p$ and the ones with tiny amounts of available space). Note that we can push $p$ as close as we want to the corner it is near and still have some such spaces, and in doing so we reduce the distance labelled $s$:
other 6 placements of 7
Also note that the bottom cube will never protrude over the plane of the edge (green) on the other side to $s$.
So we can have the reflected arrangement on the opposite side and leave a face of the unsolved cube (the left of the above picture) entirely free up to $s$ outside of it's edges adjoining the faces with $7$. On the adjacent edges to that face (top and bottom in the above picture) we may certainly place a cube entirely aligned, not causing any further constraint on that face, and on the remaining face (right in the picture above) we are free to place $2$ cubes (overlapping that last face and the single cubes placed on the two adjacent).

To place the $6$ cubes on the face only restricted by $s$ we can do this:
placing the 6
where the top cube pictured does not protrude on either side so may overhang in the region $s$ deep (it could go further on of course but the reason to make $s$ small was to provide room for the rest of the arrangement).

I do hope the explanation and pictures suffice with no real need for a geometrical argument, because I'm not sure I can give a rigorous proof.

This solution was found by Robert S. Holmes as part of a challenge called "a set of quickies" by Martin Gardner (who originally had found a solution of 20). The problem and Martin Gardner's 20-cube solution are in his Wheels, Life and other Mathematical Amusements, pub. Freeman, 1983, ch.8, p.80, 84. Gardner shows some improved solutions found by others, in that chapter's Addendum, on pp.88-93.

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    $\begingroup$ Whew, that took quite a while! $\endgroup$ – Jonathan Allan Jun 29 '16 at 1:41
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    $\begingroup$ worth it though, nice work. $\endgroup$ – Carl Jun 29 '16 at 3:10
  • $\begingroup$ Yes, you're absolutely correct.Accepted, and as someone has pointed it's no point to use Rubik's cube, I am about to ask a new related puzzle. Also, I'll be finding the post in Scientific american and M Gardner's set of puzzles :D $\endgroup$ – ABcDexter Jun 29 '16 at 14:24
  • $\begingroup$ For any position of p, it should be possible to compute the value of s; can you show that there is any position of p which enables the construction with all overlaps having finite (rather than infinitesimal) area? $\endgroup$ – supercat Jun 29 '16 at 15:08

I believe you can place

20 cubes around the center cube

You can do so by

placing 7 on the top and bottom and 6 along the sides.

Start by placing one cube on the top rotated 45 degrees. This leaves 4 open corners. You can place two cube on lets say the top open corner (Their teeny tiny corners will be on the corner, faces still touching). You can repeat this on the opposite corner. The remaining two corners can have one cube each. This adds up to 7 on the top. Repeat for the bottom.

To fill in the sides, you can place two cubes along one face, and do so for the opposite face. Similar to the top and bottom parts, you can fill in the other sides with one cube each. This puts 6 around the sides.

In total, you get 7 + 7 + 6 = 20.

  • $\begingroup$ hee hee wrong look @ answer above $\endgroup$ – Adit Kirtani Jun 30 '16 at 10:28

My answer is


This is reached by

Construct 3 layer that look like this:

      |   |   |
    |   |   |   |
      |   |   |
The middle of the middle layer is the unsolved cube. Then rotate the middle layer by 45°

  • $\begingroup$ aren't the ones touching the middle layer overlapping the ones in the bottom or top layer? $\endgroup$ – elias Jun 29 '16 at 12:11
  • $\begingroup$ @elias: No, each layer stays in its original plane when you turn the middle one by 45°. $\endgroup$ – hmakholm left over Monica Jun 29 '16 at 14:01
  • $\begingroup$ ah, ok, I get it, thanks! I thought a layer refers to one of the three layers of a single Rubik's cube. $\endgroup$ – elias Jun 29 '16 at 14:18




you can make a bed of four solved cubes, and put the unsolved one at the intersection. On the next level, you can put two on each side. Then there's only space for one each on the front and back. Then another 4-cube roof on top. 4+2+2+1+1+4 = 14


I can think of 22 by modifying one of the configuration used by @jonathan allan enter image description here

Now instead of using these 6 cubes as used by Jonathan I do not use the top one(6th cube). Now if you think carefully u can use this same configration for four sides without any interference . As this configuration obstruct's three face's and two of these faces are always the same so we get four face's with 5 cube's and 2 face's with one 5+5+5+5+1+1=22

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    $\begingroup$ So you modify a 24-cube solution in order to get a 22-cube solution -- what's the point of that? $\endgroup$ – hmakholm left over Monica Jun 29 '16 at 14:08
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    $\begingroup$ I think it is an interesting take on the problem, he generated a method there the side face is repeated 4 times, rather than twice, going to a 4 fold symmetrical solution. That in itself is interesting. On top of that scientific progress rarely goes in a straight line. Looking at new angles can both open new avenues and be illustrative. $\endgroup$ – Going hamateur Jun 29 '16 at 20:55
  • $\begingroup$ Might be useful to add that the two faces of $1$ are opposite each other and each of the $5$s has it's pictured top against an adjacent $5$'s pictured bottom pictured (thus the $1$s as shown would be left and right, each with a swastika or it's inverse when viewed face on). $\endgroup$ – Jonathan Allan Jun 29 '16 at 21:09
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    $\begingroup$ @HenningMakholm everything is not about answer , sometime's thinking process is also important . This might not be the best solution but I thought not sharing such a fantastic geometry would be a waste . $\endgroup$ – Kashish Garg Jul 2 '16 at 20:03

I'm going to start things off with



One on the bottom, one on the top
Two on a pair of opposite sides
four on the remaining pair of opposite sides



4 on bottom 4 on faces 4 on sides 4 on top (one on top to balance top 4 ---- NOT counted)

Therefore, 16.



On two opposite sides, we can get seven each.
To do this, make a grid of cubes on a table, 3 by 3.
The shift the top and bottom rows by half a cube.
Then you'll have the middle cube touching 6 others.
Remove the 2 it doesn't touch, this is the base.
Place the unsolved cube on the middle cube rotated 45 degrees, such that it touches them all.
Then place a solved cube against two of the opposite sides of the unsolved one, and two more (offset) on each of the other two sides.
That is 7+6. For the top layer, repeat the first layer.

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    $\begingroup$ hahaha nice try but try harder! hahaha! $\endgroup$ – Adit Kirtani Jun 30 '16 at 10:29
  • $\begingroup$ Yah i saw the answer above after writing it :/ $\endgroup$ – will Jun 30 '16 at 10:33

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