Von:Mike Keith (domnei@aol.comXYZXYZ) Betrifft:Re: Another number game... Newsgroups:rec.puzzles Datum:2000/08/03 >1) >>Find a 10 digit number such that the first n digits (for every 0>be divided by n and every number between 0 and 9 is used exactly once. > >2) >>If that's too easy: How many solutions are there (scfry: this might be >>another job for your "brute force" program)? > >If that's too easy: Repeat 1 and 2 for arbitrary base b. (Brute >force will get you some way, but not all the way.) > > SPOILER (partial?) . . . . . . . . . . . I've known for quite a while that these are the only solutions with base <= 25: base number (expressed in base b) 2 10 4 1230 4 3210 6 143250 6 543210 8 32541670 8 52347610 8 56743210 10 3816547290 14 9(12)3(10)5476(11)812(13)0 and I conjectured that there are no more. Your statement of the problem implies that you know the solution for all b (which I think is equivalent to proving my conjecture). If so, I would be very interested in seeing this proof. Regards, Mike Keith http://members.aol.com/s6sj7gt/mikehome.htm -------------------------------------------------------------------------------- Von:Nis Jørgensen (nis@dkik.dk) Betrifft:Re: Another number game... View this article only Newsgroups:rec.puzzles Datum:2000/08/05 On 03 Aug 2000 23:05:21 GMT, domnei@aol.comXYZXYZ (Mike Keith) wrote: >>If that's too easy: Repeat 1 and 2 for arbitrary base b. (Brute >>force will get you some way, but not all the way.) [snip partial solution] >Your statement >of the problem implies that you know the solution for all b >(which I think is equivalent to proving my conjecture). I don't think my statement implies that - it only implies that I know that brute force wont help you. >If so, I would be very interested in seeing this proof. I have a "proof" why you are _very_ unlikely to be wrong. BTW: I checked it for b<=42. It seemed like the right time to stop.