**Xi'an's Og » R**, and kindly contributed to R-bloggers)

**T**he solution to puzzle 48 given in * Le Monde* this weekend is rather direct (which makes me wonder why the solution for 6 colours is still unavailable..) Here is a quick version of the solution: Consider one column, 1 say. Since 326=5×65+1, there exists one value

*c*with at least 66 equal to

*c*. Among those (at least) 66 rows, if a pair

*(i,j)*satisfies , the problem is over. Otherwise, all are different from

*c*for those (at least) 66 rows, hence equal to one of the four remaining values. Since 65=4×16+1, for a given row

*i*in this group, there exists

*d*different from

*c*for which at least 17 are equal to

*d*. Again, either there is at least one in this group of indices, else they all are different from

*c*and

*d*, hence equal to one of the three remaining values. Then 16=3×5+1, and for a given index

*j*within this group there exists

*e*different from

*c*and

*d*for which at least 6 ‘s are equal to

*e*. Again, either there is a triplet or they all take a value different from

*c,d,e*. Since 5=2×2+1, there exists

*f*different from

*c,d,e*, for which at least 3 ‘s are equal to

*f*. Again, either end of the story or they all three take the final value

*g*, but then constitute a triplet…

**T**his week puzzle [49]: *in a lottery, 999<N<10000 tickets numbered 1,2,…,N have been sold. All those involving a 1 and a 3 to the right of the 1 are winning tickets. The percentage of winning tickets is 10%. How many tickets are there?* A manageable problem for R, obviously!

Filed under: R, Statistics Tagged: graphs, Le Monde, mathematical puzzle, R

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