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enter image description hereCan anyone explain to me the math behind the problem? I want to convert the mathematical solution into an efficient algorithm. The target sum can be any given number. For reference please check the following link out- (solution found) How to fill up the numbers in a set of empty discs drawing a pentagon?

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  • $\begingroup$ If someone is informed enough to explain to me the math it will be really helpful. Downvoting the question does not help anybody. $\endgroup$ Commented Apr 9, 2022 at 22:05
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    $\begingroup$ Does this answer your question? How to fill up the numbers in a set of empty discs drawing a pentagon? - the first answer already explains its math $\endgroup$
    – bobble
    Commented Apr 9, 2022 at 22:22
  • $\begingroup$ Where is the math? The numbers are arranged with the help of a trial and error method based on guesses. Is there a formulated way to arrange the numbers? Or do you have to cherry-pick them? Here the target sum is not the issue, the technique to arrange the numbers matters. $\endgroup$ Commented Apr 9, 2022 at 22:35
  • $\begingroup$ There may be no generic math to directly target a generic solution, but there sure may be some math to help the search for a solution. Sometimes it's just like so. Proving in this case there is no such generic math is another challenge :-) Have a nice day. $\endgroup$ Commented Apr 9, 2022 at 23:48
  • $\begingroup$ Well is it possible to solve the problem for a target sum of 10? I really need to work on the algorithm. :( $\endgroup$ Commented Apr 10, 2022 at 20:17

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Sum of all 10 numbers is $2+4+...+20=110$. Minimum sum of 5 numbers is $2+4+...+10=30$, maximum sum of 5 numbers is $110-30=80$. Let sum of angles is $2k$, then sum of side middles is $110-2k$. The sum of all sides is $2\cdot 2k+110-2k=110+2k$ (every angle participates in two sides). This number must be multiple of 5 (because every of 5 sides has equal sum), then $2k$ must be multiple of 5. Then maximum $2k$ is 80, which corresponds to sum of all sides equal $\frac{110+2k}{5}=38$.

Now we need to check existence of solution with sum of all sides equal $38$. When sum of angles is $80$ then angles must be 12, 14, 16, 18, 20 and side middles must be 2, 4, 6, 8, 10. Numbers adjacent to 2 must add up to $38-2=36$, then these numbers are 20 and 16. Numbers adjacent to 10 must add up to $38-10=28$, then these numbers are 12 and 16. Then we have sequence 20-2-16-10-12. Then two next numbers after 12 must add up to $38-12=26$, and one number must be not greater than 8 and other must be not greater than 18. Then these numbers are 8 and 18. Then we have sequence 20-2-16-10-12-8-18. Two next number must add up to $38-18=20$ and one of the numbers must be not greater than 6 and other must be not greater than 14. Then these numbers are 6 and 14. Then we have sequence 20-2-16-10-12-8-18-6-14. The only remaining number 4 closes the circle and $20+4+14=38$. This is the only possible solution (excluding reversing and shifting).

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  • $\begingroup$ I attached a picture if it helps anybody. $\endgroup$ Commented Apr 11, 2022 at 17:05
  • $\begingroup$ If we consider numbers from 1 to 10, then minimum sum of each side is 14. It can be obtained from my answer if we take minimum possible $2k=30$, calculate $\frac{110+2k}{5}=28$ and divide result by 2, as all numbers are two times less. $\endgroup$ Commented Apr 11, 2022 at 17:16
  • $\begingroup$ I am reading your comment. If I understand it correctly I can come up with the algorithm. Thank you very much for the answer. You saved the day! :D $\endgroup$ Commented Apr 11, 2022 at 17:23
  • $\begingroup$ I tried to come up with an algorithm but it is hard to do so. I need a bit more explanation regarding the sequence. Is there any sort of way we can represent this sequence through an induction method? It will be a great help. For example, we have five corners and five sides. Let corners be c and sides be s. The sequence is kinda like- 2k1+s1+2k2+s2+2k3+s3 and so on. Conditions are sn!=kn, k1!=k2...kn and s1!=s2...sn (for n gons). The summation of every three-point has to be the same and can not overlap each other. Can it be structured in a specific pattern? $\endgroup$ Commented Apr 14, 2022 at 23:35

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