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The numbers 1 to 15 should be arranged in a way that any 3 adjacent numbers' sum will be a perfect cube.

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  • $\begingroup$ The smallest sum should be 8, since it's 2's cube. And maximum sum is 27(3's cube) since the maximum possible sum can only be 42. $\endgroup$
    – Satyam
    Mar 5, 2016 at 7:38

1 Answer 1

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It is impossible. Why?

Note that the largest possible sum of any three numbers from the set is $15+14+13=42<64=4^3$ while the smallest possible sum of any three numbers will be $1+2+3=6>1=1^3$, so the only two available cubes will be $8$ and $27$

Suppose we could arrange the numbers in such a way as to have all consecutive triples adding up to be a cube. Consider what happens to a number larger than $8$.

Let $a_1,a_2,a_3,a_4$ be distinct members of the set $\{1,2,\dots,15\}$ with specifically $a_2>8$.

Suppose $a_1,a_2,a_3,a_4$ satisfies that $a_1+a_2+a_3$ is a cube and that $a_2+a_3+a_4$ is a cube. Then $a_1+a_2+a_3>8$ and is a cube, so $a_1+a_2+a_3=27$ is the only possibility. Similarly $a_2+a_3+a_4=27$ is the only possibility, but then that implies that $a_1=a_4$, a contradiction.

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  • $\begingroup$ shouldn't you take $a_2>12$, so that you are sure that there is at least one such $a_2$ since not all three of them could be on an extreme of the arrangement? $\endgroup$
    – mau
    Mar 9, 2016 at 20:21
  • $\begingroup$ @mau how does that contradict what I had already done? Sure, I left out the detail that there will always be a number greater than eight at a position other than an edge, but that should have been obvious without stating. The bound on $a_2$ could even have been relaxed to $a_2>5$, as it is only necessary to note that the sum of the triple being a cube forces what the cube is in this case. $\endgroup$
    – JMoravitz
    Mar 9, 2016 at 20:23
  • $\begingroup$ If you had stated that there is always at least one such number, I would not have commented... $\endgroup$
    – mau
    Mar 9, 2016 at 20:30

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