Le Monde puzzle [#1022 & #1023]

Another Le Monde mathematical puzzle where I could not find a solution by R programming (albeit one by cissors and papers was readily available!):

An NT is a T whose head () is made of 3 50×50 squares and whose body (|) is made of N 50×50 squares.  What is the smallest possible side of a square containing four non-intersecting NT’s when N=1,2,4? And what is the smallest value of N such that this square also contains a fifth NT?

The questions could have been solved by brute force simulation (or a knapsack algorithm?!) but I could not fathom an efficient way to code throwing T’s at random over an MxM grid.So instead I took scissors and paper and tried to fit four 1T, 2T, and 4T into the smallest squares, ending up with 4×4, 5×5, and 7×7 squares. Interestingly, four 5T also fit in a 7×7 square. And a 9×9 square accommodates the extra 7T. Compared with the  “impossible” puzzle of last week, this is pretty anticlimactic..! (Actually, once the solutions were published, I realised the square containing the T’s did not have to be with integer side. Which means the smallest square for 3Ts was incorporating the glued T’s sideway. Fortunately, this did not impact the answer for the 7T’s!)

Going back to this “impossible” puzzle, the posted solution is somewhat… puzzling in that the resolution posits that the majority rule is the optimal allocation, when I am not sure it is [optimal]. Just because, when rerunning the same R code, I found instances when the minimal acceptable number of councillors was lower than the one returned by the majority rule.

And since this post get pushed down in the queue, here is as a bonus the equally anticlimactic puzzle #1023,

Find (a) a multiplication of two three-prime-digit numbers such that all digits everywhere in the long multiplication are prime and all three intermediary products have four prime digits, while the final result has six prime digits, and (b) a multiplication of two three-digit numbers such that the digits of the first one are odd (o), the digits of the second are even (e), the three intermediary products are all of the form eoe, and the final product is of the form eoeo. [The website has two pictures to help if this description is too unclear!]

This is indeed straightforward to code with one solution to (a) and two to (b) since the number of cases to examine is quite limited.

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