**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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**Tags:** graphs, Le Monde, mathematical puzzle, R, statistics