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The setup is that we have a total of 9 persons scattered in 6 groups. Each persons having a unknown number of pens and being part of two different groups. The repartition of the groups is like so :

$$\begin{array}{|c|c|c|c|} \hline & \text{Group 1} & \text{Group 2} & \text{Group 3}\\ \hline \text{Group 4} & P_1 & P_2 & P_3 \\ \hline \text{Group 5} & P_4 & P_5 & P_6 \\ \hline \text{Group 6} & P_7 & P_8 & P_9 \\ \hline \end{array}$$

The total numbers of pens from each group are given :

$$ Group\ 1 : 204\\ Group\ 2 : 0\\ Group\ 3 : 135\\ Group\ 4 : 38\\ Group\ 5 : 93\\ Group\ 6 : 208\\ $$

The objective is to know the number of pens owned by each person.

I thought that it would obviously be a set of equations with a set of unknowns but the problem here is that we have 9 unkwnowns with only 6 equations.

$$ \begin{eqnarray} P_1 + P_4 + P_7 &= 204\\ P_2 + P_5 + P_8 &= 0 \\ P_3 + P_6 + P_9 &= 135\\ P_1 + P_2 + P_3 &= 38\\ P_4 + P_5 + P_6 &= 93\\ P_7 + P_8 + P_9 &= 208\\ \end{eqnarray} $$

Hopefully then, the second equation allows to reduce the number of unknowns to 6 but it also reduces the number of remaining equations to 5 :

$$ \begin{eqnarray} P_1 + P_4 + P_7 &= 204\\ P_3 + P_6 + P_9 &= 135\\ P_1 + P_3 &= 38\\ P_4 + P_6 &= 93\\ P_7 + P_9 &= 208\\ \end{eqnarray} $$

To top it all, those equations are not linearily independant since the first equation can be retrieved as a combination of the last four.

I am leaded to think that there might be something that I missed which could allow me to reduce the number of unkwowns even more but I can't put my hands on it.

Could anyone here help me solve this?

mqbaka mqbaka
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    There are obviously a lot of solutions that are determined by what you assign to $P_1$ and $P_4$. Do you have anything else? – Parcly Taxel Sep 30 '22 at 08:38
  • No, that's about all, though. The original problem led me to think that there should be an unique solution but I don't see what more one can find about this. – mqbaka mqbaka Sep 30 '22 at 08:49
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    You can narrow the number of solutions down somewhat, based on the constraint that all of the variables must be non-negative integers. As an exercise, I explored the problem to determine if that constraint allowed the problem-solver to deduce that there was only one unique solution. I could be mistaken. However, I don't think that there is only one unique solution, even given the constraint re non-negative integers. – user2661923 Sep 30 '22 at 09:17
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    I suggest that you re-check the original wording of the problem. If you omitted certain pertinent info, or made a typo in one of the equations, that could result in there being a unique solution. – user2661923 Sep 30 '22 at 09:20
  • @user2661923 That's exactly what I am thinking and I've been reading and re-reading the problem from top to bottom but I couldn't see what I missed. The problem is in French, I pasted it on imgur if you want to take a look https://imgur.com/a/m2ows9K – mqbaka mqbaka Sep 30 '22 at 10:10
  • Is that all you have? – Parcly Taxel Sep 30 '22 at 10:47
  • Yes that's all I have – mqbaka mqbaka Sep 30 '22 at 10:51
  • Thanks for the help guys. I think the solution would be something like what @ParclyTaxel suggested. The solution is not unique. – mqbaka mqbaka Sep 30 '22 at 11:10

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