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I do not claim credit for this puzzle; I saw it online years ago and realised it hadn't been asked here yet (as far as I could tell), so I thought I'd share it with the community here. I cannot recall the text verbatim, so I sincerely hope that I do not end up being misleading. Here goes:

You start with small and equal cubes, 27 of which are put together to form a large cube. How does one then remove 1 from the middle of the large cube to leave only 26, while not touching any of the sides of the large cube?

Xenocacia
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11 Answers11

63

Maybe this is a large cube and if you take away the middle one you get 26.



The $216$ is $6$ cubed ($6^3$) made of $27$ small cubes.  Taking away the middle '1' leaves "2 6".

Edited by asker:

Cubes in the puzzle are used in a figurative sense. Start with 27 small cubes: 2^3 = 8. Put them together to form a large cube: 8*27 = 216 = 6^3. You are now free to remove 1 from the middle and leave 26 behind, all without touching any of the sides!

Peregrine Rook
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Tom
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18

How about

Removing the centre cube via a 4th spatial dimensions in a manner analogous to removing the centre of a square comprised of 9 equal smaller squares through the 3rd dimension.

10

The center cube...

...is an ice cube, while the others are identical-in-appearance glass, ...

...so I can...

...allow the center cube to melt and leak out while I work on other puzzles...

...without touching anything.

humn
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    Well... not to be picky here, but how about the You start with small and equal cubes part? – Matsmath Sep 20 '16 at 09:26
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    @humn: like Matsmath said, the starting cubes are equal. If the centre were ice, they'd all melt together, so... nope! Sorry! – Xenocacia Sep 20 '16 at 09:28
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    You could use a laser focused on the central cube to heat up only that one ... – fffred Sep 20 '16 at 09:50
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    No problem being wrong in service of lateral-thinking variety. It can also be done purely mathematically: The "cubes" are each $8=2^3$, so 27 of them amount to the larger cube of $27 {\small\times} 8 = 216=6^3$. Simply recalculate with $216=26{\small\times} \Big(\frac{3 ,\raise.6ex{\small3}\kern-.4em\surd4}{\raise.6ex{\small3}\kern-.4em\surd{13}}\Big)\raise1.3ex{\small3}$. – humn Sep 20 '16 at 10:08
  • @Matsmath - small and equal cubes... being made of ice will not change the size (small cube). It also won't change the dimensions, so the cube might be the exact same size (equal cube, equivalent cube, equally sized cube). After all, the cubes are equal... not identical :) – Megha Sep 22 '16 at 02:22
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My answer...

You are using a 3D Modeling Software, so you just select the cube in the middle and delete it.

fdr
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The question nowhere states that

the blocks are directly connected to each other, just that they are put together to form a large cube.

So, it could also be that

the blocks are actually just nodes that are connected with sticks.

It's then fairly easy to

just cut the sticks to the cube in the center and take it out.

slvrbld
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6

Possibly

you take it out through the top. Essentially thinking of the large cube as having 4 sides a top and a bottom?

DRF
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The cube is formed floating in a vacuum with only the corners of one of the smaller cubes ever touching any other of the smaller cubes. The cube itself is half cube matter, and intermittently half cuboid empty space. A small cube in the middle of this scenario can come out of the center of the larger cube.

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Edit: Using pincer utensils to go through the 'gaps' of the cube you can avoid touching the sides. Then manipulating the cube carefully or shaking it will remove the middle cube if it is not connected at its corners, cut it out if it is.

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    I see an $11\times5$ grid, where alternate positions contain $6+5+6+5+6$ $\rm x$s. How does this relate to the question? – Peregrine Rook Sep 23 '16 at 17:23
3

As I understood

You start with more that 27 cubes

So

You can use 2 other cubes to push out the middle one

Gintas K
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  • If I understand you right, you seem to be suggesting replacing the central axis of 3 cubes with 2 cubes separated by a gap (how they stay that way is handwaved). That kinda means you remove more than 1, which is not what the question is asking for. :) – Xenocacia Sep 20 '16 at 09:43
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I will try this way...

You start with a group of 27 magnetic cubes

So...

As it's stated you can build a larger cube from there so in principle they are not built up in the larger one. Just remove 1 of those magnetic cubes from the group (implying you won't use it) and just build up your larger cube with the 26 small cubes, leaving a vacant in the middle of the structure.

Yandrakus
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  • Thanks for suggesting that, it led me to edit the question to clarify that the small cubes are put together as the large cube already. I hadn't thought that anyone would try that. – Xenocacia Sep 20 '16 at 09:56
  • Well it was a bare try while thinking about more possibilities :P – Yandrakus Sep 20 '16 at 09:57
  • By glueing and orienting the cubes strategically, the middle one can actually push out the center of one side and move into its place as soon as you stop touching that side. – humn Sep 20 '16 at 10:06
3

How about

the literal product 1 X 1 X 1 ........X 1 such that there are 27 1's multiplied together. This would result in a cube(1). You could remove one the central 1 leaving 26 cubes (1) without touching the sides?

sriram
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A possibility is:

Put a hole to the cube that will be placed in the middle.Then after arranging all the cubes, keep your finger into the hole of the middle cube and take it away.

Another possibility:

Put some adhesive to your finger, then touch the middle cube(after arrangement) and keep it some time to make strong contact with the cube. After that take away the cube out.

KSR
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