Three logical people, A, B, and C, are wearing hats with positive integers painted on them. Each person sees the other two numbers, but not her own, and each person knows that the numbers are positive integers and that one of them is the sum of the other two.
They take turns (A, then B, then C, then A, etc.) in a contest to see who can be the first to determine her number.
During round one, A, B, and C pass. In round two, A and B again pass, at which point C states that she now knows all three numbers and that their sum is 144.
How did C figure this out?
(In reply to
re: Possible solution by pcbouhid)
It is given that one of the numbers is the sum of the other two.
Without looking at other possible solutions, let us assume that the numbers are as Harry posted: (18, 54, 72).
C would see 18 and 54, and therefore would logically surmise the three numbers to be (18, 54, 36) or (18, 54, 72). If her number was 36, C knows B would logically deduce the three numbers to be (18, 18, 36) or (18, 54, 36). Yet if B's were 18, B would know C would know on round one the three numbers (18, 18, 36); but C passed on round 1, therefore B would know the three numbers to be (18, 54, 36) on round two; yet, B passed on round 2, thus C knows by deduction that her own number must be 72, and by simple math knows the total of the three numbers (18, 54, 72) to be equal to 144.
|
Posted by Dej Mar
on 2009-05-03 21:03:33 |