Dick Hess: The Logical Hats Puzzle. Logicians A, B and C each wear a hat with a positive integer on it such that the number on one hat is the sum of the numbers on the other two. They can see the numbers on the other two hats but not their own. They are given this information and asked in turn if they can identify their number. In the first round A, B and C each in turn say they don't know. In the second round A is first to go and states his number is 50. What numbers are on B and C?
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Answers Below
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Hi Ed
I was delighted to see an old puzzle of mine doing the rounds (Logical Hats Puzzle). This was originally published in the Sunday Times magazine as puzzle number 1814, and was reprinted in a collection of these puzzles, Brainteasers by Victor Bryant in 2002 - highly recommended if you like tough puzzles.
Whenever I see the puzzle again I wonder whether it was best to choose 50 as A's number. Some people (mistakenly) think that the first question to A is unnecessary and the problem simplified without it. It is possible to have B get the answer 26 on the second round of questions and still have unique numbers for A and C, or perhaps a second round of 'no's followed by A deducing 68. But, on reflection, I'm happy with 50 since it is the simplest possible expression of the problem whilst still requiring an appreciation of the general method in order to solve it - making it harder doesn't really involve any further insights in solving.
Of course, any triplet of numbers satisfying a+b=c can eventually be deduced, and in this general context the puzzle has relationships with fibonacci numbers (no suprise) and Euclid's method of finding the GCD of two numbers.
Regards
Jonathan Welton
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A has 50, B has 20, and C has 30.
Seeing B and C, A knows he has either 10 or 50. Assume he has 10 - we'll prove he doesn't.
C sees A with 10 and B with 20. C knows he has either 10 or 30. If he has 10, B will know for certain he has 20, since all numbers are positive (non-zero). But B says he doesn't know his number, so C knows he has 30.
So if A has 10, either B or C will certainly know their own number. Since they both say they DON'T know their numbers, A knows he doesn't have 10, so he can say he certainly has 50.
Ken Duisenberg
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B has 20 and C has 30.
Note that if two people have the same number, then the third person immediately knows his number is the sum of those two, because the alternative, that he has the difference of those two, is impossible, since 0 is not positive. For any other combination of numbers, A cannot determine whether his number is the sum or difference of two numbers in the first round.
So, the only ways for B to know what his number is when A did not are when the two numbers B sees are the same, and they are in the ratio 2:1 with the larger number on A. In the latter case, if B has the difference, or the same as C, then A would have known immediately when he saw two identical numbers, so he knows A:B:C are in ratio 2:3:1.
The reasoning continues in this fashion, each turn the logician being able to know what his number is only when the other alternative would have given it away to another logician previously. So when C goes, he knows his hat is neither the same as A's, nor the same as B's, nor half A's. He determines his number if A and B are identical, or if the ratio A:B:C is 1:2:3, 2:1:3, or 2:3:5.
So when A goes again, he knows his number is not equal to or twice either of the others, not half of B's, and not in a 2:3 ratio with B's. He determines his number when A:B:C are in any of these ratios: 3:2:1, 3:1:2, 4:3:1, 4:1:3, 5:2:3, 8:3:5. (In each case, the alternative is a combination somebody else would have already solved.) Out of these, only 5:2:3 allows A's number to be 50.
It is also interesting to note that the logician with the largest number always figures it out first. The initial cases where one logician can figure out his number without other information are ones where he has the largest number. The other deductions a logician can make come from concluding that some other logician did not have the largest number and thus he himself does not have the difference between the other two numbers, and hence must have the sum, or largest number. Thus, all the deductions remain in this form, and the logician with the largest number always figures it out first.
Joseph DeVincentis
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Dear Ed,
The key to solving the logical hats puzzles is as follows:
Each person knows that the number on his hat is either the sum or the difference of the other two. He also knows his number is positive.
Therefore with no other knowledge, A can only know the number on his hat if B and C are equal. Therefore A's "no" response rules out the numbers 2k,k,k.
Similarly B's "no" rules out k,2k,k and 2k,3k,k.
C's "no" rules out k,k,2k and 2k,k,3k and k,2k,3k and 2k,3k,5k.
A would say "no" unless the situation were 3k,2k,k or 4k,3k,k or 3k,k,2k or 4k,k,3k or 5k,2k,3k or 8k,3k,5k.
Therefore since A says yes, it must be one of those situations. The only one where A can have 50 is if B has 20 and C has 30.
Luke Pebody
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The numbers on A, B, and C’s hats are 50, 20, and 30, respectively. When A looks at the hats of B and C, he realizes that his number is either 10 or 50. Well, what would happen if it was 10? In this case, B would look at the hats of A and C and not know whether his number was 40 or 20. Then, C would look at the hats of A and B and know that his number was either 30 or 10. But wait! If C’s number was 10, then B would have known that his number was 20 (10, 0, 10 is not a possibility). So, in fact, C would know with certainty that his number was 30.
But none of this happened. Thus A realizes that his number must be 50, not 10.
Daniel Scher , Brooke Précil
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Here is the answer to the logical hats puzzle:
B is 20 and C is 30 (or vice versa)
A sees 20 and 30 doesn't know if he's wearing 10 or 50.
B sees 30 and 50 doesn't know if he's wearing 20 or 80
C sees 20 and 50 doesn't know if he's wearing 30 or 70
A now knows he's not 10 because if B saw 10 and 20 he'd know his hat is 30.
-Andy Ofiesh
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Hi Ed,
Here's the reasoning behind the problem: In the first round A sees 20 on B and 30 on C and cannot tell if he has 50 or 10. B sees 30 and 50 and cannot tell between 20 and 80. C sees 20 and 50 and cannot tell between 30 and 70. In the second round A considers that he may have 10 but realizes that if he has 10 then C would have seen 10 and 20, giving rise in C's mind to possibilities 10 or 30 for C. But C would reason that if he has 10 then B would have seen 10 and 10 and know his number. Thus A reasons that C would have known his number on the first round if A had 10. He can thus eliminate 10 and knows he has 50.
All the best, Dick Hess
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Logician B's hat says 20. Logician C's hat says 30.
A realizes that his number is either 10 or 50. Since he doesn't have enough information the first round, he passes.
If A is wearing a 10, then C will see 10 and 20. He will conclude that he is wearing 10 or 30. If B saw two 10's, he would realize that his own hat says 20. Since this doesn't happen, C would realize that his own hat says 30, if A's hat was a 10. Since C also passes, A's hat cannot be a 10, therefore he realizes that his own hat says 50.
John Boozer IV
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Hello,
In my humble opinion, the solution is:
A = 50, B = 20, C = 30
Proof:
First, a lemma - if a logician sees two identical numbers K on the other two hats, they immediately concludes that his number is 2*K (since it can not be 0).
At the beginning of the first round, A reasons as follows: "I see 20 and 30 thus my number is either 50 or 10". Thus he says he doesn't know.
At the beginning of the second round, A reasons as follows: "My number is either 50 or 10". Let's assume my number is 10. In that case, C saw A= 10 and B = 20. C must have reasoned this way: "My number is either 30 or 10. If it was 10, then B would have seen 10 and 10 and he would have announced his number. But B didn't say anything thus my number must be 30" (end of A reasoning about C reasoning)."
Logician A continues to reason: "However, C didn't say anything. The only conjecture is that my assumption that my number is 10 is incorrect and my number is 50."
Sincerely,
Vladimir Valenta
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When A says 50, B has 20, and C has 30.
Rod Bogart
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Hi Ed,
The Logical Hats Puzzle answer: A:50, B:20, C:30
Solution:
Let's first consider the Case 1: A:x, B:2x, C:x
where X - any positive integer.
A thinks: "I have x or 3x (no way to tell)"
B thinks: "The only possibility is that I have 2x"
B states that his number is 2x
Now let's procees to the Case 2: of A:x B:2x C:3x
A thinks: "I have either x or 5x (no way to tell)"
B thinks: "I have either 2x or 4x (no way to tell)"
C thinks: "I have either x or 3x. If that's x, we'd have the Case 1 and B
would state his number. therefore, it's not x but 3x."
C states his number is 3x
Now, to the final Case 3: A:5x, B:2x, C:3x
In the first round, no one can tell what the numbers are. But in the second round, A thinks: "I have either x or 5x. If I have x, it's Case 2 (A:x B:2x C:3x) and C would state his number. However, he didn't, which means I have not x, but 5x." A states his numbner is 5x
That's the answer. In our case, A states his number is 50, so x = 10 and B:20 C:30
Thanks,
Igor
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Ha! Found the solution to logical hats puzzle: B=20, C=30. Any case
with 5x,2x,3x works. Similarly, 3x,x,2x works. Nice puzzle! Thanks,
John Elton