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Logicians, hats, and numbers (Posted on 2006-08-14) Difficulty: 5 of 5
Adam, Bob, and Chuck, three perfectly intelligent logicians, are sitting facing each other with a hat on each of their heads so that each can see the others' hats but they cannot see their own. Each hat, they are told, has a (non-zero) positive integer on it, and the number on one hat is the sum of the numbers on the other two hats. The following conversation ensues:

Adam: I do not know the number on my hat.
Bob: I do not know the number on my hat.
Chuck: I do not know the number on my hat.
Adam: I do not know the number on my hat.
Bob: I do not know the number on my hat.
Chuck: I do not know the number on my hat.
Adam: I do not know the number on my hat.
Bob: I do not know the number on my hat.
Chuck: I do not know the number on my hat.
Adam: The number on my hat is 1691.

Adam was correct. What are the numbers on the other two hats?

No Solution Yet Submitted by Avin    
Rating: 3.7778 (9 votes)

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Solution Perfectly Intelligent | Comment 7 of 20 |
(In reply to Help needed. by Vernon Lewis)

It basically just means they all think alike and are intelligent enough not to miss any information or say they don't know when they could. Of course if they can say "I don't know" at any time, it wouldn't make much sense to report on what they said.

I was going to use logic similar to JLo's logic, but it would seem obvious that it couldn't be "267 178 89" if it was really "1691 356 1335", because none of the numbers match up! So my guess is there is something not quite right in that logic.


The way to approach a problem like this (where you can rule out possiblities by concluding they don't know) is to figure out what the numbers would be if they DID know. The only information they know, is that their numbers are all positive. Thus, if somebody saw two of the same numbers, they would know their number couldn't be 0, and thus it must be the sum of those numbers.

So for the case, 4, 6, 2, B knows his number is 6 because if it was 2, A would have figured out his number sooner. Thus:

A: Possiblities 4 or 8 Response: Don't know
B: Possibilties 6 (A's possiblities 4 or 8) or 2 (A's possibilities: 4) Response: 6

Thus it would seem this could be used recursively like JLo's post said. For example, with 4, 6, 10: (Note that C's other possiblity, 2, is shown in the above example.)

A: Possibilities are 4 or 16 Response: Don't know
B: Possiblities are 6 (A: 4 or 16) or 14 (A: 4 or 24) Response: Don't know
C: Possibilities are 10 (B: 6 (A: 4 or 16) or 14 (A: 4 or 24)) or 2 (B: 6 (A: 4 or 8 ) or 2 (A: 4)): Response: 10

Since both people before C didn't know, we must remove any path which anyone only has one choice. Thus C:2,B:2 only has one choice, A:4, and then C:2 only has one other choice, B:6.

For 3 "I don't know" use 16, 6, 10

A: 4 or 16 Response: I don't know
B: 6 (A: 4 or A:16) or 26 (A: 16 or A:36) Response: I don't know
C: 10 (B: 6 (A: 4 or A:16) or B:26 (A: 16 or A:36)) or 22 (B:6 (A: 16 or 28) or 38 (A: 28 or 48)) Response: I don't know
A: 16 (C: 10 (B: 6 (A: 4 or A:16) or B:26 (A: 16 or A:36)) or C:22 (B:6 (A: 16 or A:28) or B:38 (A: 28 or A:48))) or 4 (C: 10 (B: 6 (A: 4 or A:16) or B:14 (A: 4 or A:24)) or C:2 (B: 6 (A: 4 or A:8 ) or B:2 (A: 4))) Response: 16

Notice that what A thinks he has doesn't change from his first reponse to his second. The only difference is he has more to think about since if his number was 4, three "I don't know" responses wouldn't have happened.

So to figure out what numbers A saw, a chain needs to occur with 9 "I don't know" parts. If we start with for 0 "I don't know" responses 2,1,1, we can see for 1 "I don't know" response a solution is 2, 3, 1 (since B knows it can't be 2,1,1) and then for 2 "I don't know" responses, could be 2, 3, 5 (since C knows it can't be 2, 3, 1) and so on to 9 "I don't know" responses we reach A having 89, B having 34, and C having 55.

Now after all this computation, we are in luck, because it just so happens that 89 is a factor of 1691. The other factor is 19, so all that is needed is to multiply 55 and 34 by 19 to get the numbers they had, which are 1045 for Chuck and 646 for Bob.

  Posted by Gamer on 2006-08-16 01:25:55
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