Sum three terms & digits

[Peter Pein]

Hi Max,

 sorry to say, but you've got a typo in the last line of numbers:
> 14854, 48310, 199058, 374854
instead of
14854, 48310, 159058, 414854, 1648310, 5159058, 15414854, 15748410, 31058958,
175414854, 415748410, 1631058958.

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.

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
>>
> 
>