Enigma chain thread

[palou]

Angelus said:

palou said:

20 and 30 is sadly incorrect.

 

No it's not.

If two people have the same number, then the third person immediately knows his number is the sum of those two, because the alternative, that he has the difference of those two, is impossible, since 0 is not positive. For any other combination of numbers, A cannot determine whether his number is the sum or difference of two numbers in the first round.

So, the only ways for B to know what his number is when A did not are when the two numbers B sees are the same, and they are in the ratio 2:1 with the larger number on A. In the latter case, if B has the difference, or the same as C, then A would have known immediately when he saw two identical numbers, so he knows A:B:C are in ratio 2:3:1.

The reasoning continues in this fashion, each turn the logician being able to know what his number is only when the other alternative would have given it away to another logician previously. So when C goes, he knows his hat is neither the same as A's, nor the same as B's, nor half A's. He determines his number if A and B are identical, or if the ratio A:B:C is 1:2:3, 2:1:3, or 2:3:5.

So when A goes again, he knows his number is not equal to or twice either of the others, not half of B's, and not in a 2:3 ratio with B's. He determines his number when A:B:C are in any of these ratios: 3:2:1, 3:1:2, 4:3:1, 4:1:3, 5:2:3, 8:3:5. (In each case, the alternative is a combination somebody else would have already solved.) Out of these, only 5:2:3 allows A's number to be 50.

Sorry about that. I should have redone the problem before posting.