Le Monde puzzle [48: resolution]

December 4, 2010

(This article was first published on Xi'an's Og » R, and kindly contributed to R-bloggers)

The 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 $a_{1i}$ equal to c. Among those (at least) 66 rows, if a pair (i,j) satisfies $a_{ij}=c$ , the problem is over. Otherwise, all $a_{ij}$ 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 $a_{ij}$ are equal to d. Again, either there is at least one $a_{jk}=d$ 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 $a_{ij}$ ‘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 $a_{ij}$ ‘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…

This 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