Arthur and Bert each writes down a positive integer on a piece of paper and then shows it to Charles. Charles then writes two numbers on a blackboard, visible to Arthur and Bert: one of them is the sum of Arthur's and Bert's numbers, and the other is a random number.
After this Charles asks Arthur if he knows Bert's number. If Arthur says he doesn't know, then he asks Bert if he knows Arthur's number. If Bert says he doesn't know, Charles continues with Arthur, then if necessary with Bert and so on... until he gets a positive answer.
When will Charles get a positive answer?
(In reply to
X = 13, Y = 16 by Steve Herman)
I may not have been clear. If X = 13 and Y = 16, we will never
get more than 4 nos. I suspect that that is because 13 / (16-13)
is less than 5, but I haven't proved that yet.
If A = 4 and B = 9, then A knows from the start that B has 9 or 12, and
B knows from the start that A has 4 or 7. B's second no, which
places him in the 4 to 9 range, is all A needs to hear.
If A had 2 and B had 11, then A knows from the start that B has 11 or
14, and B knows from the start that A has 2 or 5. B's first no is
all that A needs to hear. A will answer yes on the 2nd round,
because anybody in the room knows that B is between 4 and 12.
In summary, If X = 13 and Y = 16,
A will answer yes on round one if he has 13, 14, 15
Otherwise, B will answer yes on round 1 with 1, 2, 3, 13, 14,15
Otherwise, A will answer yes on round 2 with 1, 2, 3, 10, 11, 12
Otherwise, B will answer yes on round 2 with 4, 5, 6, 10,11, 12
Otherwise, A will answer yes on round 3 with 4, 5, 6, 7, 8, 9
re: X = 13, Y = 16 (continued)
