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# What Question Must Be Asked?

## Question

Three logicians, Carroll, Kurt and Kleene, are captured by the evil villainous mastermind, Moriarty. They are put in adjacent cells each of which contains a number of coins. All of them can count the number of coins in their own cells, but not in anyone else’s. They are told that each cell has at least one coin, and at most nine coins, and no two cells have the same number of coins. The logicians must use their skills of deductive reasoning to escape their cells. The three of them will ask Moriarty a single (yes or no) question each, which he will answer truthfully ‘Yes’ or ‘No’. Every one hears the questions and the answers. Moriarty will free the logicians only if one of them correctly works out the total number of coins in all three cells. Here’s how the conversation between them ensues.

Carroll: Is the total number of coins an even number?

Moriarty: No

Kurt: Is the total number of coins a prime number?

Moriarty: No.

If Kleene has ﬁve coins in his cell, what question should he ask Moriarty in order to ensure that at least one of the logicians work out the total number of coins in the cells?

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Kleene: Is the total value of the coins 15?

There are currently 3 possibilities for their total: 9, 15, and 21.  If the answer to the question is yes, everybody knows the total.  If the answer to the question is no, one two of the three (both Carroll and Kurt) should be able to deduce whether the total is 9 or 21.

If anyone has 7, 8, or 9 coins, they can deduce that there must be 21 coins (they would have to total more than 9).
If anyone has 1, 2, or 3 coins, they can deduce that there must be 9 coins (they couldn't sum to 21 given the rules).

Since Kleene has 5 coins, there is no way to make 21 coins without someone having 7 and the other having 9.  There's no way to make 9 coins without someone having 1 and the other having 3.

Edited by Molly Mae

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What will happen if either Carroll or Kurt has 4 or 6 coins in their cell when Kleene's question on the "total  = 15" is NO? One can still arrive at a total 9 or 21. Carroll and Kurt don't know that Kleene has 5.

One can possibly figure out a question to pose wrt the "total number of coins", but what if the OP stated "number of coins in each cell"? Then the answer is obviously 9 or 21 since no one would be able to figure out how to get to a total of 15 (four cases). So I think that, in such an instance, one should merely ask ... "Does the sum of the coins have an integer square root?" YES would mean a total of 9, NO would mean a total of 21.

Sorry for side tracking!

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9 = 1+5+3 (if YES)

21= 7+5+9 (if NO)

But we won't know who's got 1 and 3 versus 7 and 9!

Edited by rocdocmac
expansion

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10 hours ago, rocdocmac said:
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What will happen if either Carroll or Kurt has 4 or 6 coins in their cell when Kleene's question on the "total  = 15" is NO? One can still arrive at a total 9 or 21. Carroll and Kurt don't know that Kleene has 5.

One can possibly figure out a question to pose wrt the "total number of coins", but what if the OP stated "number of coins in each cell"? Then the answer is obviously 9 or 21 since no one would be able to figure out how to get to a total of 15 (four cases). So I think that, in such an instance, one should merely ask ... "Does the sum of the coins have an integer square root?" YES would mean a total of 9, NO would mean a total of 21.

Sorry for side tracking!

Since Kleene does have 5, though, if the answer to the question is no and the total isn't 15, Carroll and Kurt can't have 4 or 6 coins.  They must have either 1 and 3 or 7 and 9.

Anyone who has 1 or 3 can deduce that they won't have enough coins for 21.  Anyone who has 7 or 9 can deduce that they have too many for 9.

If the answer isn't 15, 4 and 6 coins aren't possible since Kleene has 5.

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Of course you're right!

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Just a small correction.

Spoiler

No one can have 8 coins nor 2.

If the total is 21: 5+7+9 is the only possibility.
If the total is 4: 1+3+5 is the only possibility

Edited by harey
Tried to hide it. Did not really succeed.

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7 minutes ago, harey said:
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Just a small correction.

Hide contents

No one can have 8 coins nor 2.

If the total is 21: 5+7+9 is the only possibility.
If the total is 4: 1+3+5 is the only possibility

Correct.  I clarified that at the end of the post.

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Good work straight to the point ...I screwed around and came up with the same Q ,but a lot longer getting there

Kleene: Is the total value of the coins 15?

There are currently 3 possibilities for their total: 9, 15, and 21.  If the answer to the question is yes, everybody knows the total.  If the answer to the question is no, one two of the three (both Carroll and Kurt) should be able to deduce whether the total is 9 or 21.

If anyone has 7, 8, or 9 coins, they can deduce that there must be 21 coins (they would have to total more than 9).
If anyone has 1, 2, or 3 coins, they can deduce that there must be 9 coins (they couldn't sum to 21 given the rules).

Since Kleene has 5 coins, there is no way to make 21 coins without someone having 7 and the other having 9.  There's no way to make 9 coins without someone having 1 and the other having 3.

Small correction ;  for 21 no 8 possible ;  The same for 9 i.e. no 2 possible

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Haha, you're the third person to correct me on that point.

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Do we all agree on what the task really is?

is it (1) ensure someone  can correctly state the total number of coins —seems like that is solved by Molly Mae, OR (2) ensure someone can correctly identify the number of coins in each cell?

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Assuming we need to find the number in each cell, here’s a variant that will give the others a bit more information: “is the total either 9 or 21?”

this shows the others that Kleene has 4, 5, or 6. I don’t see that that’s enough yet. But maybe y’all do.

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Better still...

”is the smaller of the number of coins of my companions in the set {1,2,3,4} AND the larger in the set {6,7,8,9}?”

the only person who has that precise dilemma is the one with 5.

if the answer is YES, then the others know the sum is 15, and can subtract to get each others’ numbers.

if the answer is NO, then they know the sum is either 9 or 21, and, per previous discussion, each one can tell which, and can identify the other’s number

Thank you, Devoe, for the impossible question, and Molly Mae, rocdocmac, harey, and Donald Cartmill, for showing that more information is available.

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Proof of my claim...

Here is a terse list of all the cases deemed possible by a person with i coins, for i in [1,9]. Possible sums are listed, each with the only possible pairs of numbers of coins.

1: 9(35,26), 15(68,59)

2: 9(34,16), 15(67,58,49)

3: 9(24,15), 15(57,48)

4: 9(23), 15(56,38,29), 21(89)

5: 9(13), 15(46,37,28,19), 21(79)

6: 9(12), 15(45,27,18), 21(78)

7: 15(35,26), 21(68,59)

8: 15(34,25,16), 21(67,49)

9: 15(15,24), 21(57,48)

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On 2/8/2018 at 8:32 PM, Molly Mae said:

Correct.  I clarified that at the end of the post.

I was so stuck in my solution that is exactly the same as yours except that the exclusion came earlier that I reacted too quickly.

Next time, I will read more carefully, promised

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