Sum three terms & digits

[Max Alekseyev]

On 5/23/07, Peter Pein <petsie at dordos.net> wrote:

> 14854, 48310, 159058, 414854, 1648310, 5159058, 15414854, 15748410, 31058958,
> 175414854, 415748410, 1631058958.

I corrected my program again, and also got this sequence.
Bingo! ;)

> The last two integers contain all digits except "2". The sum of the next
> integer n_imposiible and 415748410 + 1631058958 must end with a "2". So the
> last digit of n_impossible has to be a "4", which is allready in 415748410.

Agree.

Max







>
> Cheers,
> Peter
>
> Max Alekseyev schrieb:
> > Peter,
> >
> > Actually, there was a bug in my program. Thanks for your program and
> > the sequence.
> > After fixing the bug, I have got the following sequence (with one
> > extra term as compared to your sequence):
> >
> > 1, 2, 3, 4, 5, 7, 8, 9, 10, 14, 15, 31, 32, 33, 34, 35, 37, 38, 39,
> > 40, 43, 44, 63, 64, 65, 68, 69, 73, 76, 79, 80, 83, 86, 88, 96, 116,
> > 118, 119, 120, 124, 125, 128, 140, 267, 426, 440, 445, 446, 447, 460,
> > 474, 604, 733, 774, 775, 777, 778, 779, 785, 797, 818, 819, 830, 873,
> > 888, 889, 890, 893, 894, 913, 915, 916, 939, 945, 977, 1111, 1114,
> > 1128, 1148, 1224, 1227, 1229, 1400, 2704, 2729, 2732, 2939, 2940,
> > 2972, 2973, 4223, 4320, 6608, 6623, 6680, 6688, 7743, 7760, 9608,
> > 14854, 48310, 199058, 374854
> >
> > And this sequence is finite and complete!
> >
> > Proof:
> > Suppose that the next term is x, then the sum
> > s = 199058 + 374854 + x
> > may contain only decimal digits 2 and 6.
> > First note that if s deliver the required number x, then so does the
> > number s'=(s mod 10^6).
> > Therefore, we can limit our search for s to 6-digit numbers, but they
> > give no solution.
> >
> > Max
> >
> > On 5/23/07, Peter Pein <petsie at dordos.net> wrote:
> >
> >>  I think you misunderstood Eric's intention (or I did so). He wrote:
> >> "9+10+14=33 and "33" shares nothing...". Therefore the union of the
> >> lhs-digits
> >> shall have an empty intersection with the rhs-digits (10 and 14 would
> >> share
> >> the "1").
> >>
> >> My Mathematica-approach is:
> >>
> >> noCommonDigit[{a_, b_}] :=
> >>   Module[{c = b + 1, abDigits = IntegerDigits[{a, b}]},
> >>    While[
> >>     Intersection[Flatten[{abDigits, IntegerDigits[c]}],
> >>                  IntegerDigits[a + b + c]] =!= {},
> >>     c++];
> >>    c]
> >>
> >> Prepend[
> >>   Last /@ NestList[Through[{Last, noCommonDigit}[#1]] & , {1, 2}, 100],
> >> 1]
> >>
> >> {1, 2, 3, 4, 5, 7, 8, 9, 10, 14, 15, 31, 32, 33, 34, 35, 37, 38, 39,
> >> 40, 43,
> >> 44, 63, 64, 65, 68, 69, 73, 76, 79, 80, 83, 86, 88, 96, 116, 118, 119,
> >> 120,
> >> 124, 125, 128, 140, 267, 426, 440, 445, 446, 447, 460, 474, 604, 733,
> >> 774,
> >> 775, 777, 778, 779, 785, 797, 818, 819, 830, 873, 888, 889, 890, 893,
> >> 894,
> >> 913, 915, 916, 939, 945, 977, 1111, 1114, 1128, 1148, 1224, 1227,
> >> 1229, 1400,
> >> 2704, 2729, 2732, 2939, 2940, 2972, 2973, 4223, 4320, 6608, 6623,
> >> 6680, 6688,
> >> 7743, 7760, 9608, 14854, 48310, 159058}
> >>
> >>
> >> Regards,
> >> Peter
> >>
> >
> >
>
>