In a political party, there are as many cells as there are members and each member belongs to at least one cell. Each cell has five members and an arbitrary pair of cells only shares one member. How many members are there in this political party?
Back to the mathematical puzzles of Le Monde (science leaflet of the weekend edition)! In addition to a tribune by Cédric Villani celebrating the 100th anniversary of the death of Henri Poincaré, this issue [now of last week] offers this interesting challenge. So much interesting that I could only solve it for three (instead of five) members and could not see the mathematical notion behind the puzzle…
Let us denote by n the number of both the cells and the number of members. Then, when picking an arbitrary order on the sets, if ij denotes the number of members in set j already seen in sets with lower indices, we have the following equality on the total number of members
and the constraints are that i2<2, i3<3, i4<4, i5<5, and ij<6, for j>5. Hence, i2+i3+i4+i5+…+in≤5n-15, which implies n≥15.
Now, in terms of analytics, I could not go much further and thus turned to an R code to see if I could find a solution by brute force. Here is my code (where the argument a is the number of elements in each set): Continue reading