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Coasterfrenzy

Harder hats on deathrow version

Question

Three prisoners in deathrow are put together. They each get a unique number on their forehead. They are NOT allowed to speak to each other, only to their guard. They only get 2 clues:
- The number on their head is always higher than zero
- One of the numbers is the total of the other two numbers.

If they can guess their numbers, they are free to go.

- Person 1 gets asked if he knows his number. He says he doesn't.
- Person 2 gets asked the same, but also replies that he doesn't know
- Person 3 is being asked but doesn't know either.

Since none of them answered wrong. They get another chance.

- Person 1 again doesn't know.
- Person 2 also doesn't know.
- Person 3 however does know the answer this time: It's 148 on his head. This answer is correct and he is allowed to leave.

Question: What are the numbers on the heads of person 1 and 2 and how does person 3 know his own number?

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I assume that they also all get the clue that their numbers are unique.


the other numbers are 37 and 3*37= 111

Here’s the argument:


If the numbers are n, 2n, and 3n, then the person with 3n would recognize that fact because seeing n and 2n, the only choices would be n and 3n, but since n would not be unique, it would have to be 3n.

So the numbers are not n, 2n, and 3n.

Therefore, if someone sees n and 3n, they know they don’t have 2n, or the 3n person would have recognized it. So they must have 4n.

So the person with 148 could be 4n, seeing 37 and 111.

Neither of the other two could rule out their two possibilities.


 

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On 4/16/2020 at 10:33 PM, Coasterfrenzy said:

Three prisoners in deathrow are put together. They each get a unique number on their forehead. They are NOT allowed to speak to each other, only to their guard. They only get 2 clues:
- The number on their head is always higher than zero
- One of the numbers is the total of the other two numbers.

If they can guess their numbers, they are free to go.

- Person 1 gets asked if he knows his number. He says he doesn't.
- Person 2 gets asked the same, but also replies that he doesn't know
- Person 3 is being asked but doesn't know either.

Since none of them answered wrong. They get another chance.

- Person 1 again doesn't know.
- Person 2 also doesn't know.
- Person 3 however does know the answer this time: It's 148 on his head. This answer is correct and he is allowed to leave.

Question: What are the numbers on the heads of person 1 and 2 and how does person 3 know his own number?

Prisoner 3  sees the other two have  both the same number  on their head  -  which is 74  but Prisoners 1 & 2 only see number 148 and 74 on the other two prisoners heads respectively.  So they cannot really tell if they have to subtract or add to get their own number.

Correct?

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21 hours ago, qubits said:

Prisoner 3  sees the other two have  both the same number  on their head  -  which is 74  but Prisoners 1 & 2 only see number 148 and 74 on the other two prisoners heads respectively.  So they cannot really tell if they have to subtract or add to get their own number.

Correct?

Unfortunately no.

- The numbers on their heads are unique.
-  Prisoner 3 only knows his number when questioned the 2nd time.

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Nice beginning... (If this were the full solution, first round would be enough.)

Oh, I got it now!!! Great!!!

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Still confused.

Spoiler

Say the numbers are 37 111 148.

On the end of the first round C thinks:

- I can have 111-37=74 or 111+37=148.

- If the guy with 111 sees 37 and 74, he announces 37+74=111 (as he knows he cannot have 74-37=37).

- As he did not announce 111, I only can have 148.

So why the second round?

 

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