I found the beginning of a reference : a paper published in 1992 by Stephen Davis, I should be able to get hold of it tomorrow, The 1987 paper by Mendelsohn also seems to address the problem….

]]>The solution goes like this: *“The optimal solution must correspond to settings where rows include 3 or 4 entries, because using 5 entries eliminates too many cases from the following rows. Here is a grid with 32 entries (2 rows of 4 and 8 rows of 3). A simple reasoning shows that having more than 2 rows with 4 entries imposes corresponding rows with 2 entries.”* Hardly convincing… The last part is wrong: in my 34 entries example, I get 4 rows with 4 entries and 6 with 3 entries. None with 2.

(It is the first equation listed, where k would equal 2.) k of 2 gives a solution of 10*sqrt(10) + 10 + 10 = 51 (or 52, depending if the floor or ceiling function is used, but I’m pretty sure it’s the floor function.)

So, Le Monde is almost certainly wrong.

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