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Yakk · 04-26-2004, 10:42 AM

Assumption 1: Any conclusion you can reach while knowing none of the numbers is valid dispite any knowledge of some of the numbers.

So, lets add a person who doesn't get to see any of the numbers. Anything this person can deduce is ALSO true for anyone else.

Let X be someone who knows none of the numbers. Note that A B and C know what X knows.

Lets work out what X knows:
A passes.
X knows that if B==C, then A would not pass. Thus, X knows B!=C.
B passes.
X knows that if A==C, then B would not pass. Thus, X knows A!=C.
X knows that if A==2C, then either (A=B+C => B==C) or (B=A+C).
But, X knows that B!=C, so, if A==2C, then B=A+C. B would know A and C and hence know B, but B passed.
So, X knows A!=2C.
C passes
Simularly to above, X now knows A!=2B.
In addition, if (A==3B) then either (A==B+C=>C==2B) or (C=A+B). But, X knows C!=2B, so if A==3B then C=A+B. Thus, if A==3B, C would know C=4B. C passed.
So, X knows A!=3B.
A now goes.
X whispers in A's ear what he knows.
Suppose B was 10 and C was 40. Then, A entertains two possibilities.

Either A=B+C or C=A+B.

If C=A+B, then 40=A+10 => A=30. Then, A = 3B. But, X knows that A!=3B.
So, if B is 10 and C is 40, then A knows A=50.

The only possiblity that would make the above not work is if B or C could figure out the answer before A went.

B goes. B sees 40 and 50. B knows he is either 10 or 90.
If B was 90, then A would have seen 40 and 90. This doesn't help, A would have passed.
If B was 10, then A would have seen 40 and 10. This doesn't help, A would have passed.
C goes. C sees 50 and 10. C knows he is either 40 or 60.
If C was 60, then B would have seen 60 and 50. Then B would have thought he was either 10 or 110. If C was 60 and B thought he was 10, then C would have thought that B would have thought that A would have seen 10 and 60. A would have passed, no problem.
If C was 60 and B thought he was 110, then B would have thought A would have seen 60 and 110. A would have passed, which is consistent.
If C was 40, then B would have seen 50 and 40. Then B would have thought he was either 10 or 90. If C thought that B thought he was either 10 or 90, then B passing is consistent.
So, C cannot deduce what his number is, and passes.

Thus, both C and B would have passed, and X can tell A enough that A can deduce his number. Hence, A knows his number.

04-26-2004, 10:42 AM	#9 (permalink)
Yakk Wehret Den Anfängen! Location: Ontario, Canada	Assumption 1: Any conclusion you can reach while knowing none of the numbers is valid dispite any knowledge of some of the numbers. So, lets add a person who doesn't get to see any of the numbers. Anything this person can deduce is ALSO true for anyone else. Let X be someone who knows none of the numbers. Note that A B and C know what X knows. Lets work out what X knows: A passes. X knows that if B==C, then A would not pass. Thus, X knows B!=C. B passes. X knows that if A==C, then B would not pass. Thus, X knows A!=C. X knows that if A==2C, then either (A=B+C => B==C) or (B=A+C). But, X knows that B!=C, so, if A==2C, then B=A+C. B would know A and C and hence know B, but B passed. So, X knows A!=2C. C passes Simularly to above, X now knows A!=2B. In addition, if (A==3B) then either (A==B+C=>C==2B) or (C=A+B). But, X knows C!=2B, so if A==3B then C=A+B. Thus, if A==3B, C would know C=4B. C passed. So, X knows A!=3B. A now goes. X whispers in A's ear what he knows. Suppose B was 10 and C was 40. Then, A entertains two possibilities. Either A=B+C or C=A+B. If C=A+B, then 40=A+10 => A=30. Then, A = 3B. But, X knows that A!=3B. So, if B is 10 and C is 40, then A knows A=50. The only possiblity that would make the above not work is if B or C could figure out the answer before A went. B goes. B sees 40 and 50. B knows he is either 10 or 90. If B was 90, then A would have seen 40 and 90. This doesn't help, A would have passed. If B was 10, then A would have seen 40 and 10. This doesn't help, A would have passed. C goes. C sees 50 and 10. C knows he is either 40 or 60. If C was 60, then B would have seen 60 and 50. Then B would have thought he was either 10 or 110. If C was 60 and B thought he was 10, then C would have thought that B would have thought that A would have seen 10 and 60. A would have passed, no problem. If C was 60 and B thought he was 110, then B would have thought A would have seen 60 and 110. A would have passed, which is consistent. If C was 40, then B would have seen 50 and 40. Then B would have thought he was either 10 or 90. If C thought that B thought he was either 10 or 90, then B passing is consistent. So, C cannot deduce what his number is, and passes. Thus, both C and B would have passed, and X can tell A enough that A can deduce his number. Hence, A knows his number. __________________ Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.