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?

(In reply to Perfectly Intelligent by Gamer)

Nice problem!

Following on from Gamer's description, here's the full expansion of all the ratios for each round where the logician could determine his number

At each stage, you take all the ratios from the previous 2 rounds, and sum the other two positions. E.g. in round 2, (2,1,1) from round 1 is changed to (2,3,1) (B's 3 is the sum of A'2 plus C's 1.) For the first 3 rounds, permutations of (2,1,1) are added, for the trivial cases where the other two values are the same.

1A: (2,1,1) - (i.e. B and C are the same, A is the sum)
2B: (1,2,1), (2,3,1)
3C: (1,1,2), (2,1,3), (1,2,3), (2,3,5)

4A: (3,2,1), (4,3,1), (3,1,2), (4,1,3), (5,2,3), (8,3,5)

5B: (1,3,2), (2,5,3), (1,4,3), (2,7,5), (3,4,1), (4,5,1), (3,5,2),(4,7,3), (5,8,3),(8,13,5)

6C: (3,2,5), (4,3,7), (3,1,4), (4,1,5), (5,2,7), (8,3,11), (1,3,4), (2,5,7), (1,4,5), (2,7,9), (3,4,7), (4,5,9), (3,5,8),(4,7,11), (5,8,13),(8,13,21)

7A: (5,3,2), (8,5,3), (7,4,3), (12,7,5), (5,4,1), (6,5,1), (7,5,2), (10,7,3), (11,8,3), (18,13,5), (7,2,5), (10,3,7), (5,1,4), (6,1,5), (9,2,7), (14,3,11), (7,3,4), (12,5,7), (9,4,5), (18,7,9), (11,4,7), (14,5,9), (13,5,8), (18,7,11), (21,8,13), (34,13,21)

8B: (3,8,5), (4,11,7), (3,7,4), (4,9,5), (5,12,7), (8,19,11), (1,5,4), (2,9,7), (1,6,5), (2,11,9), (3,10,7), (4,13,9), (3,11,8), (4,15,11), (5,18,13), (8,29,21), (5,7,2), (8,11,3), (7,10,3), (12,17,5), (5,6,1), (6,7,1), (7,9,2), (10, 13, 3), (11,14,3), (18,23,5), (7,12,5), (10,17,7), (5,9,4), (6,11,5), (9,16,7), (14,25,11), (7,11,4), (12,19,7), (9,14,5), (18,27,9), (11,18,7), (14,23,9), (13,21,8), (18,29,11), (21,34,13), (34,55,21)

[...uh, I forum software window is too small for me to see previous lines and expand the series. Cut&Paste also doesn't work in FF.]

You eventually get many possible ratios. After testing all those ratios, only (89, 34, 55) produces integral values when A=1691 (89,34,55) x 19 = (1691,646,1045).

Posted by Mat on 2006-08-16 13:19:29