Raul, Looking at the original source of the problem, the first question has only 9 solutions, not 10. So your solution is correct. Here's the original problem statement and solution: https://xianblog.wordpress.com/2019/06/18/le-monde-puzzle-1104/ Here's my version of the solution
pal=.]#~(=|.&.":&>) coins=:pal}.i.1e3 (}.i.1e3)-.coins,,+/~1e3(>#])coins 21 32 43 54 65 76 87 98 201 So only 9 answers. Skip Cave On Wed, Jun 19, 2019 at 2:56 PM Raul Miller <[email protected]> wrote: > I came back to think about this and I noticed a mistake: > > On Wed, Jun 19, 2019 at 10:12 AM Raul Miller <[email protected]> > wrote: > > I don't know what the tenth value would be. (If I remove the "1" coin, > > I get 18 values that can't be paid with three coins.) > > That should have been: > > ... that can't be paid with less than three coins. > > I suppose it's possible that the tenth value might be zero? But then > the problem statement would have been "that can't be paid with one or > two coins" and we would have an implicit constraint against a "0" > coin. I do have that implicit constraint in my implementation in this > thread, but... I can't quite make sense of 0 being the tenth value we > couldn't pay with less than three coins... > > Not sure if this is of any help, though. > > -- > Raul > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
