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### #1

Posted 30 March 2008 - 07:30 AM

He explains to them he has written a positive integer on each hat, and

that one of the numbers is the sum of the other two. Each logician can

see only the numbers on the other two hats.

A prize is offered to the first person able to be certain of the number on

his own hat. The wizard starts questioning the logicians in order, starting

over again if none of them can be certain of his number.

There is no guessing.

Each logician must answer: "My number is ___" or "I don't know."

[1] Can any of the logicians win the prize?

[2] If so, which one?

[3] How many rounds of questions will it take?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #2

Posted 30 March 2008 - 08:15 AM

### #3

Posted 30 March 2008 - 08:41 PM

Think harder ... one of them can.I don't think they can ever be sure. Whatever they see, and whatever the others say, their number could still be the sum of the other two numbers, or one of the smaller ones.

How?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #4

Posted 30 March 2008 - 09:05 PM

### #5

Posted 30 March 2008 - 09:10 PM

### #6

Posted 30 March 2008 - 09:58 PM

### #7

Posted 30 March 2008 - 10:03 PM

You guys are missing something obvious

Spoiler for answer

You're right. And as I can see this is the only case where someone would be able to tell. And it would be in the first round.

### #8

Posted 31 March 2008 - 02:17 AM

Call the logicians Fred, George, Harry, and assume they are questioned in that order.

Assume the integers are a>b>=c. [a=b>c is not possible.]

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

### #9

Posted 31 March 2008 - 04:06 AM

Spoilers anyone?

Call the logicians Fred, George, Harry, and assume they are questioned in that order.

Assume the integers are a>b>=c. [a=b>c is not possible.]Spoiler for Say you're FredSpoiler for Say you're GeorgeSpoiler for etc

How do you know what numbers you are looking at? You said say you are fred and see a and c. But how do you know you are looking at a and c. You could be looking at a and b.

In your second one, how do you know you see b and c. You could be looking at a and b.

### #10

Posted 31 March 2008 - 04:21 AM

You can assume three numbers and ask what each logician in turn thinks and knows.How do you know what numbers you are looking at? You said say you are fred and see a and c. But how do you know you are looking at a and c. You could be looking at a and b.

In your second one, how do you know you see b and c. You could be looking at a and b.

Then you can substitute variables [a, b, c] and do the same.

To get started, try this single question.

You're the first to be questioned.

You see 5 and 13.

What do you know about your number?

*The greatest challenge to any thinker is stating the problem in a way that will allow a solution.*

- Bertrand Russell

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