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Is it possible to split the set of 14 consecutive cubes $1^3,2^3,\ldots,14^3$ into two subsets of equal sums?

There has to be a more efficient approach than brute force, right? Because with brute force, you'd have to try a ridiculous amount of combinations... I posted a similar question a few days earlier, found here: Sums of Consecutive Cubes (Trouble Interpreting Question)

However, this problem is much harder, or at least has many more combinations to try, because there is no longer the condition that the two subsets have to have the same number of elements.

I would appreciate any and all help.

Thanks

Edit: Originally, I wanted to deal with the first 13 consecutive cubes, but as André Nicolas pointed out, it would not be possible given that the first 13 cubes sum to an odd value.

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  • $\begingroup$ You may want to start with an even cube. $\endgroup$ Commented Apr 1, 2015 at 2:43

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Edit: The original question was about $1^3+\cdots+13^3$. The new question is about $1^3+\cdots +14^3$. Precisely the same argument works.

Old answer: It cannot be done, since the sum $1^3+\cdots+13^3$ is odd. That is clear, we are adding an odd number of odd numbers and some evens.

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  • $\begingroup$ We have 6 evens and 7 odds. An odd sum can be achieved with one set containing 3 evens and 3 odds, and the other set containing 3 evens and 4 odds, for example. $\endgroup$ Commented Apr 1, 2015 at 1:17
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    $\begingroup$ If we could split into two sets with equal sums $S$, then the total sum would have to be $2S$, which is even. But the total sum is not even. $\endgroup$ Commented Apr 1, 2015 at 1:20
  • $\begingroup$ You're right. Sorry I overlooked such a simple fact. I've made an edit to the problem to allow for a more interesting scenario. $\endgroup$ Commented Apr 1, 2015 at 1:29

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