bonanova 84 Posted March 30, 2008 Report Share Posted March 30, 2008 A wizard selects three excellent logicians and places hats on their heads. 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? Quote Link to post Share on other sites

0 Guest Posted March 30, 2008 Report Share Posted March 30, 2008 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. Quote Link to post Share on other sites

0 bonanova 84 Posted March 30, 2008 Author Report Share Posted March 30, 2008 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. Think harder ... one of them can. How? Quote Link to post Share on other sites

0 Guest Posted March 30, 2008 Report Share Posted March 30, 2008 After more thinking I still stand by my original answer that you cannot know. If you see two numbers yours could either be the sum of the two you see, or yours could equal the larger number minus the smaller number you see. Quote Link to post Share on other sites

0 Guest Posted March 30, 2008 Report Share Posted March 30, 2008 I agree with Noct... it is impossible to determine... Quote Link to post Share on other sites

0 Guest Posted March 30, 2008 Report Share Posted March 30, 2008 You guys are missing something obvious He never said they had to be DIFFERENT integers...which means that if two of the hats have the same number on them, the third person will know what his number is...in which case he would know in the first round. Quote Link to post Share on other sites

0 Guest Posted March 30, 2008 Report Share Posted March 30, 2008 You guys are missing something obvious He never said they had to be DIFFERENT integers...which means that if two of the hats have the same number on them, the third person will know what his number is...in which case he would know in the first round. 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. Quote Link to post Share on other sites

0 bonanova 84 Posted March 31, 2008 Author Report Share Posted March 31, 2008 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.]Say you are Fred, and you see a and c. What do you know about your own number?Say you are George. You hear Fred say I don't know my number. You see b and c. What do you know about your number?You get the idea ... Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 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.]Say you are Fred, and you see a and c. What do you know about your own number?Say you are George. You hear Fred say I don't know my number. You see b and c. What do you know about your number?You get the idea ... 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. Quote Link to post Share on other sites

0 bonanova 84 Posted March 31, 2008 Author Report Share Posted March 31, 2008 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. You can assume three numbers and ask what each logician in turn thinks and knows. 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? Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 You can assume three numbers and ask what each logician in turn thinks and knows. 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? Could either be 8 or 18 Quote Link to post Share on other sites

0 bonanova 84 Posted March 31, 2008 Author Report Share Posted March 31, 2008 Could either be 8 or 18 Yes. And you answer "I don't know my number." Next step: What does that tell the next logician. What numbers might he see. What does that permit him to conclude. Can you take this to the place where one logician eventually knows his number? Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 Yes. And you answer "I don't know my number." Next step: What does that tell the next logician. What numbers might he see. What does that permit him to conclude. Can you take this to the place where one logician eventually knows his number? No it would just continue infinitely. If you have an 8, he will see 8 and 5, and you will have said you don't know. He could either have 13, or 3. Or if you have 18, he will see 5 and 18, so he will know he either has 13, or 23. So he will have to say he doesn't know. Then it will come to the third person, and he will see 13 and your number. If it's 13 and 8, he will know he either has 5, or 21 so he won't know. If he sees 13 and 18, he will either have 5, or 31. So he will have to say he doesn't know. And it will be back to you and you will see 5 and 13, and will still not know your number. Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 Here is what I think you are going for: a is me, b is 5 and c is 13. I could have 8 or 18, so I don't know. b sees 8/13 or 18/13 and knows he can have 5/21 for my 8, or 5/31 for my 18. Giving him 5, 21 and 31 to pick from, so he doesn't know. c sees 8/5 or 18/5 and knows he can have 3/13 for my 8, or 13/23 for my 18. He also takes into account what b is thinking, and adds into the mix 13/29, 3/39, 23/39, and 13/49. Giving him 3, 13, 23, 29, 39, and 49 to pick from, so he doesn't know. Since I realize that the number of possibilities will continue to increase indefinitely, I concentrate on the lowest possible number and work from there. If c thinks he can possibly have a 3, and b has a 5 (also being the smallest he believes possible for himself) then my number could be a 2 or an 8, so I don't know. b starts to think the same way, and realizes that I think I could have a 2, and c thinks he could have a 3, making his number a 1 or a 5. He doesn't know. c, also realizing it, gets a 2 from me and a 1 from b, giving him a 1 or a 3. He doesn't know. Me, now, I get left with a 1 and a 1, making mine a 2 if everyone really had those low number, but alas, they do not, so I don't really know. b, hearing that I don't know, knows that the whole "lowest number" thing didn't pan out. I guess this could continue until you exhausted the possibilities up the ladder of numbers, but I'm not really seeing how... Could you give a bigger hint? Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 Here is what I think you are going for: a is me, b is 5 and c is 13. I could have 8 or 18, so I don't know. b sees 8/13 or 18/13 and knows he can have 5/21 for my 8, or 5/31 for my 18. Giving him 5, 21 and 31 to pick from, so he doesn't know. c sees 8/5 or 18/5 and knows he can have 3/13 for my 8, or 13/23 for my 18. He also takes into account what b is thinking, and adds into the mix 13/29, 3/39, 23/39, and 13/49. Giving him 3, 13, 23, 29, 39, and 49 to pick from, so he doesn't know. Since I realize that the number of possibilities will continue to increase indefinitely, I concentrate on the lowest possible number and work from there. If c thinks he can possibly have a 3, and b has a 5 (also being the smallest he believes possible for himself) then my number could be a 2 or an 8, so I don't know. b starts to think the same way, and realizes that I think I could have a 2, and c thinks he could have a 3, making his number a 1 or a 5. He doesn't know. c, also realizing it, gets a 2 from me and a 1 from b, giving him a 1 or a 3. He doesn't know. Me, now, I get left with a 1 and a 1, making mine a 2 if everyone really had those low number, but alas, they do not, so I don't really know. b, hearing that I don't know, knows that the whole "lowest number" thing didn't pan out. I guess this could continue until you exhausted the possibilities up the ladder of numbers, but I'm not really seeing how... Could you give a bigger hint? You have some logic off. there could only be two possible choices for numbers. b can never have that many choices. His is either the compliment of the smaller number he sees, or the sum of both of them. Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 You have some logic off. there could only be two possible choices for numbers. b can never have that many choices. His is either the compliment of the smaller number he sees, or the sum of both of them. From MY point of view, he could have the 3 options, because I don't know what my number is. I'm saying that, being all logicians, they think the same, so each knows that the others are contemplating what my numbers are, as well as the others. Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 oops i meant c not b, sorry. Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 oops i meant c not b, sorry. I agree, it gets pretty convoluted after b. What I was saying is that if b is taking into account what I am thinking about my own numbers, and c does this for both me and b, you get the added possible numbers. Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 I agree, it gets pretty convoluted after b. What I was saying is that if b is taking into account what I am thinking about my own numbers, and c does this for both me and b, you get the added possible numbers. But c shouldn't have that many possibilities, just the same amount of possibilities as b (in your eyes). Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 But c shouldn't have that many possibilities, just the same amount of possibilities as b (in your eyes). c isn't just taking into account that I don't know my number, he is taking into account that neither of us do. Don't get me wrong, I still don't see how this could be solvable (not saying it's not possible, I just don't see it), I'm just saying that you could end up with an infinite list by continually thinking about what the others are thinking because they don't know their number. That list of numbers isn't what c actually thinks his number is, it is what the OTHERS could see as possibly what c thinks is his number, given that nobody knows their own number. Now my head hurts... I'll try to check back on this one tomorrow... Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 (edited) GOT IT BONA!!!I came to result but i am not sure...if anyone can find the flaw please mention The three numbers are- x, y, z=x+y (assuming unequal, if it is equal...it is too easy) and x>y so, z>x>y now let A's hat has x, B's hat has y and C's hat has z Case1: 1. Ask A, Ans: "I dont know", because his numbers may be z+y or z-y(as z>y, all positive integer) 2. Ask B, as he will see z>x, so, considering A's answer, the z+y=x is not possible, only z-y=x is possible. So, he can tell his number by subtracting x from z (z-x) . THE RESULT IS OBTAINED by B. Case2: 1. Ask B, Ans: "I dont know", because his numbers may be z+x or z-x(as z>x, all positive integer) 2. Ask C, as he will see x>y, so, considering B's answer, the z+x=y is not possible, only z-x=y is possible. So, he can tell his number by subtracting y from x (x-y) . THE RESULT IS OBTAINED by C. Case3: 1. Ask C, Ans: "I dont know", because his numbers may be x+y or x-y(as x>y, all positive integer) 2. Ask A, as he will see z>y, so, considering B's answer, the x-y=z is not possible, only x+y=z is possible. So, he can tell his number by subtracting y from z (z-y) . THE RESULT IS OBTAINED by A. So, THE SECOND PERSON TO ASK CAN LOGICALLY SAY HIS NUMBER ON HIS OWN HAT....IF HE IS ENOUGH INTELLIGENT...LIKE ME Edited March 31, 2008 by storm Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 GOT IT BONA!!!I came to result but i am not sure...if anyone can find the flaw please mention The three numbers are- x, y, z=x+y (assuming unequal, if it is equal...it is too easy) and x>y so, z>x>y now let A's hat has x, B's hat has y and C's hat has z Case1: 1. Ask A, Ans: "I dont know", because his numbers may be z+y or z-y(as z>y, all positive integer) 2. Ask B, as he will see z>x, so, considering A's answer, the z+y=x is not possible, only z-y=x is possible. So, he can tell his number by subtracting x from z (z-x) . THE RESULT IS OBTAINED by B. Case2: 1. Ask B, Ans: "I dont know", because his numbers may be z+x or z-x(as z>x, all positive integer) 2. Ask C, as he will see x>y, so, considering B's answer, the z+x=y is not possible, only z-x=y is possible. So, he can tell his number by subtracting y from x (x-y) . THE RESULT IS OBTAINED by C. Case3: 1. Ask C, Ans: "I dont know", because his numbers may be x+y or x-y(as x>y, all positive integer) 2. Ask A, as he will see z>y, so, considering B's answer, the x-y=z is not possible, only x+y=z is possible. So, he can tell his number by subtracting y from z (z-y) . THE RESULT IS OBTAINED by A. So, THE SECOND PERSON TO ASK CAN LOGICALLY SAY HIS NUMBER ON HIS OWN HAT....IF HE IS ENOUGH INTELLIGENT...LIKE ME I'm having trouble understanding your proof. Is there anyway you could spoiler it using some example numbers, instead of x,y,z please? Quote Link to post Share on other sites

0 Guest Posted March 31, 2008 Report Share Posted March 31, 2008 (edited) I'm having trouble understanding your proof. Is there anyway you could spoiler it using some example numbers, instead of x,y,z please? There is a problem in my previous post on the second case...it should be x+y in the second line instead of x-y...and adding in place of subtracting.... came to result but i am not sure...if anyone can find the flaw please mention Let, the three numbers are- x=5, y=3, z=x+y=8 (assuming unequal, if it is equal...it is too easy) and x>y so, z>x>y now let A's hat has 5(x), B's hat has 3(y) and C's hat has 8(z) Case1: 1. Ask A, he will see 3 and 8. Ans: "I dont know", because his numbers may be 8+3 or 8-3 (all positive integer) 2. Ask B, as he will see 8 and 5, so, considering A's answer, the 8+y=5 is not possible, only 8-y=5 is possible. So, he can tell his number by subtracting 5 from 8 (8-5=3) . THE RESULT IS OBTAINED by B. Case2: 1. Ask B, he will see 5 and 8 Ans: "I dont know", because his numbers may be 8+5 or 8-5(all positive integer) 2. Ask C, as he will see 5 and 3, so, considering B's answer, the z+5=3 is not possible, only z-5=3 is possible. So, he can tell his number by adding 3 to 5 (3+5=8) . THE RESULT IS OBTAINED by C. Case3: 1. Ask C, he will see 5 and 3. Ans: "I dont know", because his numbers may be 5+3 or 5-3(as x>y, all positive integer) 2. Ask A, as he will see 3 and 8, so, considering B's answer, the x-3=8 is not possible, only x+3=8 is possible. So, he can tell his number by subtracting 3 from 8 (8-3=5) . THE RESULT IS OBTAINED by A. So, THE SECOND PERSON TO ASK CAN LOGICALLY SAY HIS NUMBER ON HIS OWN HAT....I think it explains Edited March 31, 2008 by storm Quote Link to post Share on other sites

0 bonanova 84 Posted April 1, 2008 Author Report Share Posted April 1, 2008 Case1: 1. Ask A, Ans: "I dont know", because his numbers may be z+y or z-y(as z>y, all positive integer) 2. Ask B, as he will see z>x, so, considering A's answer, the z+y=x is not possible, only z-y=x is possible. So, he can tell his number by subtracting x from z (z-x) . THE RESULT IS OBTAINED by B. You don't say why y=x+z is not possible. [equivalently, y-z=x] That is, if x=5, y=3, z=8, A sees 3, 8 and knows he's 5 [z-y] or 11 [z+y]. B sees 5, 8 and knows he's 3 [z-x] or 13 [z+x]. If B were 13, A would have seen 13 and 8 and would know he's 5 or 21 and say I don't know. B would see 5 and 8 and know he's 3 or 13. How then does B know his number? Quote Link to post Share on other sites

0 Guest Posted April 2, 2008 Report Share Posted April 2, 2008 Case1: 1. Ask A, Ans: "I dont know", because his numbers may be z+y or z-y(as z>y, all positive integer) 2. Ask B, as he will see z>x, so, considering A's answer, the z+y=x is not possible, only z-y=x is possible. So, he can tell his number by subtracting x from z (z-x) . THE RESULT IS OBTAINED by B. You don't say why y=x+z is not possible. [equivalently, y-z=x] That is, if x=5, y=3, z=8, A sees 3, 8 and knows he's 5 [z-y] or 11 [z+y]. B sees 5, 8 and knows he's 3 [z-x] or 13 [z+x]. If B were 13, A would have seen 13 and 8 and would know he's 5 or 21 and say I don't know. B would see 5 and 8 and know he's 3 or 13. How then does B know his number? I got your point....i that case i think it is unsolvable...actually i thought they know who has the added result... ...i'm stumped!! Quote Link to post Share on other sites

0 Guest Posted April 2, 2008 Report Share Posted April 2, 2008 I got your point....i that case i think it is unsolvable...actually i thought they know who has the added result... ...i'm stumped!! I agree I think it is impossible to know your number. Quote Link to post Share on other sites

## Question

## bonanova 84

A wizard selects three excellent logicians and places hats on their heads.

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?

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