Suppose the sum of the volumes of $n$ cubes is 1. Then no matter what $n$ is I need to prove they can be put inside a cube of volume $\leq 2$ such that they do not overlap.

I am totally going nuts on how to approach this one.

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    $\begingroup$ Try the $2$-dimensional version of this problem first. Essentially, you order them from big to small, put the biggest ones in first on the bottom, then stack them on a new "level" once you run out of room on the bottom, then repeat this until done. Some clever algebra shows this always work. $\endgroup$ – Potato Sep 10 '13 at 2:31
  • $\begingroup$ The 2-dimensional version came up at math.stackexchange.com/questions/109280/…? $\endgroup$ – Gerry Myerson Sep 15 '13 at 4:47
  • $\begingroup$ I don't see that this can be done for $n=2$. How are you going to fit two cubes of side 79 inside a cube of side 126? $\endgroup$ – John Bentin Oct 3 '13 at 19:17
  • $\begingroup$ @BrianTrial: It is the volumes, not the surface areas, that must sum to $1$. Also, the volume of a cube of side $2$ is $8$, which is not $\leqslant2.$. $\endgroup$ – John Bentin Mar 18 '18 at 9:58
  • $\begingroup$ Thanks John, I realize my mistake now. Perhaps it could still be an interesting problem restated: What is the smallest volume of a cube that can contain all the other cubes no matter what n is? $\endgroup$ – Brian Trial Mar 18 '18 at 16:29

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