Another Logic Puzzle

[durron597]

Three perfect logicians are each given a number stuck to their forehead. They are told that all three numbers are distinct positive integers such that two of them add to the third.

Logician A looks at B and C and says "I do not know what my number is."

Logician B looks at A and C and says "I do not know what my number is."

Logician C looks at A and B and says "Then my number must be 50".

Logician A and B then both say, "Ah ha, my number must be _____", where "____" is different for the two logicians (obviously).

1) How did C figure out his number?
2) What are A and B's numbers?
3) How did they figure out their numbers?

[SheetWise]

<font color="white">
1) If A and B had the same number, and they had to be positive integers, then his was the sum.
2)25 and 25.
3)The same way.
</font>

[durron597]

Did you see the word "distinct" in the OP?

[kyro]

Ah, I knew that wasn't the answer but couldn't figure out why. I know the answer but only because I've seen it before. Now I'll try to prove it on my own.

[Jman28]

<font color="white"> A's number is 20
B's number is 30

C knows that he either has 50 or 10.

When A looks at 30 and 50, he could have either 80 or 20. He doesn't know.

Had he seen 30 and 10, he still wouldn't know (40 or 20).

Then B looks at 20 and 50, and doesn't know his number. It could be 70 or 30.

Had he looked instead at 20 and 10, he would know that his number was 30, since he couldn't duplicate the 10. This is how C knows that he couldn't possibly have 10.

Therefore, he must have 50.

Now, A looks at 50 and 30 and knows that he must have 20, because this number allows C to figure out his own, while 80 would not.

B does the same, sorta.

Right? Badly explained?

</font>

[robokop]

This is probably wrong but...

<font color="white">Since A can't figure out his number, it means b isn't either 0.5c or 2c. (Since if it was, he would know he must be the sum of the two numbers, as his number must be distinct, and positive).

Since B can't figure out his number, it means the same, but for a and c.

So C knows this, and sees the numbers 20 and 30. This means he must be 10 (30-20), 50 (30+20) or -10 (20-30). But he can't be negative. He also can't be 10, as then the guy seeing 10 and 20 would know he was 30 for the reasons outlined above.

Therefore, C knows his number must be 50.</font>

[durron597]

This is correct, but you didn't explain how A &amp; B figure out their numbers.

Also can you generalize? What would happen if A's number was 136, B's number 221, and C's number 357?

[Warik]

I might be wrong, but I don't think there's enough information here for Logician C to be able to say "my number must be 50" - assuming of course he really is a "perfect" logician.

Why?

Assume A = 20, B = 30, and C = 50.

C KNOWS FOR A FACT that A's number is 20 and B's number is 30.

C also KNOWS FOR A FACT that his number is thus either 50 (30 + 20 = 50) or 10 (20 + 10 = 30).

Another poster said that C would know that he was 50 because if B saw 20 and 10, then he'd know that he was 30 because you can't duplicate the 10. Ok... fine... but who says B didn't see 20 and 11? B could see 20 and 11 and then not know if he was 9 or 31. C doesn't know what's on his forehead, and I'm assuming the 3 individuals are not allowed to share information. C has no idea what B saw, and thus can't act on that information.

So... B knows that he sees 20 and 50... and he KNOWS that since he doesn't see one of the two perfect sets of numbers (numbers which must be either added or subtracted to arrive at the correct result because the other operation would result in duplication), he can't guess his own number... however, C does not know which number he has on his forehead, and so he can't use B's lack of ability to arrive at a conclusion to determine his own number, since as far as "C" knows, 50 and 10 are just two of the infinite possibilities "C" could have written on his forehead.

Now... I don't have an answer for a possible solution to this puzzle... just pointing out that I think the solutions presented already (which are acting on the above principle) are fallible.

I think more information is required in the problem statement to arrive at a conclusion here.

Either that or I've had too much to drink and have made an ass of myself.

Let me know either way. thx.

[GrekeHaus]

[ QUOTE ]
I might be wrong, but I don't think there's enough information here for Logician C to be able to say "my number must be 50" - assuming of course he really is a "perfect" logician.

[/ QUOTE ]

This is part of the information that we're given to help solve the problem. The fact that he was able to conclude this helps us solve the problem.

[ QUOTE ]
Another poster said that C would know that he was 50 because if B saw 20 and 10, then he'd know that he was 30 because you can't duplicate the 10. Ok... fine... but who says B didn't see 20 and 11? B could see 20 and 11 and then not know if he was 9 or 31. C doesn't know what's on his forehead, and I'm assuming the 3 individuals are not allowed to share information. C has no idea what B saw, and thus can't act on that information.

[/ QUOTE ]

Suppose A has 20 and C has 11. We have two possibilities: Either B has 9 or he has 31.

Case 1:
A=20
B=9
C=11

There's no way for A or B to deduce what their number is. And when it's C's turn to act, there's no way for him to determine whether he has 11 or 29. Since C can't determine this, we know that (20,9,11) isn't a possible solution.

Case 2:
A=20
B=31
C=11

Same thing here. Nobody is able to deduce what number they have, so it's not a possible solution.

Most integer solutions won't work for this reason. In order for C to be able to say for sure what his number is, he has to have the information that either A or B would have been able to guess their numbers had they been something else.

I've seen this problem before, but I'm yet to see a proof of the uniqueness of this solution. So your questions are valid.

[sirio11]

First, suppose a person Z sees numbers X and Y where Y &gt; X

Then Z has 2 options X+Y and Y-X

The only way to Z to do not know is if X,Y,Y-X and X+Y are different.

X,Y,X+Y are clearly different and Y&gt;Y-X, so the only possibility to be equal is Y-X = X this is Y = 2X

So, if Y=2X Z knows and if Y is different from 2X
Z does not know.

Since A and B do not know then C knows that his number is different from 2A,2B,(1/2)A,(1/2)B

The 2 options for C are A+B and B-A (wlog B&gt;A)

Since C can find his number, then 1 of these options should be equal to 2A,2B,(1/2)A,(1/2)B.

It's easy to check that A+B cannot be equal to any of these options. So, B-A should be equal to 1 of these.

2B=B-A # since 2B&gt;B-A

(1/2)B=B-A implies A = (3/2)B # since B&gt;A

2A=B-A implies B=3A

but since C=A+B=50 then 4A=50 # since A is an integer

so the only option left is (1/2)A=B-A

then B = (3/2)A

Now C=A+B=50 therefore (5/2)A=50 and A=20

then B = (3/2)20 = 30

and the solution is unique.