There are three logicians each of whom has a positive integer painted
on their forhead. Two of the integers sum to the third (all the
logicians are aware of this fact and can see all the integers except
their own). Each logician is asked if they know what number is on
their head. The first answers no, as does the second and third. Upon
repeating the proccess, the first answers, "yes my number is 50". What
were the other 2 integers?
spoiler
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The other numbers are 20 and 30.
Let the three numbers be x, y, z.
The first person saying no must mean both of y+z, yz are possible: ie
y!=z.
The second person saying no must mean both of x+z, xz are possible:
ie x!=z, and x!=2z.
The third person saying no must mean both of x+y, xy are possible:
So x!=y, and if x>y then:
xy != x or x/2, so x!=2y
if x<y then:
yx != x or x/2, so x!=y/2 or 2y/3
Finally, the first person says 50. So since y+z is always possible (y+z
can't equal z, 2z, y, 2y, y/2, 2y/3 without contradicting one of the
above), yz must be impossible, and x = y+z.
If y>z, we must have yz = z or 2z or y or 2y or y/2 or 2y/3, ie y=2z or
y=3z, ie x = 3z or z = 4z.
If y<z, we must have zy = z or 2z or y or 2y or y/2 or 2y/3, ie z=2y or
z=3y or z=3y/2 or z=5y/3, ie x = 3y or 4y or 5y/2 or 8y/3.
Since 50 is not a multiple of 3, 4, or 8, we must have x = 5y/2, ie y =
20, and z = xy = 30.
Stephen
PARTIAL
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If the first guy had seen 25 and 25, he'd have known the answer
immediately, since 25+(some positive number on his forehead) couldn't
possibly equal 25. But he didn't realize it until the second goround,
so I would guess that the other two numbers were 24 and 26. Not sure
how to prove it, though.
Arthur
I would venture to say 25 and 25, although he should've known the first
time around. My other guess was 50 and 100, but then whoever had 100
would've known faster.
Pb
Spoiler:
Call them A, B, C is order.
If the numbers are in proportion (2, 1, 1) (will always be given in order
A, B, C), and reduced by the GCF), then A could have given his number the
first time; in any other case, A could not have.
Similarly, if the numbers are (1, 2, 1), B could have given his numbers
the first time. If the numbers are (2, 3, 1), then B could deduce that
the numbers were (2, 1, 1) or (2, 3, 1), and eliminate the former by A's
silence.
Continuing in this manner:
A1: (2, 1, 1)
B1: (1, 2, 1), (2, 3, 1)
C1: (1, 1, 2), (2, 1, 3), (1, 2, 3), (2, 3, 5)
A1: (3, 1, 1), (4, 3, 1), (3, 1, 2), (4, 1, 3), (5, 2, 3), (8, 3, 5).
Since 50 is not divisible by 3, 4, or 8, the only possibility is that they
are in proportion 5, 2, 3;
B has 20; C has 30.


Mark Jeffrey Tilford
til...@ugcs.caltech.edu
Firstly, no two numbers can be equal otherwise the third one would have
told his number in the first go itself.
hence..starting with all the possibilites...step by step..considering
from A's point of view..
A >
(3,1,2)
(3,2,1)
(4,3,1)
(4,1,3)
(5,2,3)
(5,3,2)
(5,4,1)
(5,1,4)
If A says, his no. is 50, then (B, C) = (20,30),(30,20),(40,10),(10,40)
Well, but its certainly not obvious, how come A got to know that his
number is 50? And no one answered it here... (probably, A cheated)
Nishu

Why does X stand for a kiss?
Well, both me and Milk Tilford both answered it, actually. In all of the
cases other than (20,30), one of B or C would have known their number by
the time it got around to them. (For example, C doesn't just know all
number are different; he also knows that the difference between A and Bs
numbers isn't equal to one of A or Bs numbers).
Stephen
Nishu wrote:
> Well, but its certainly not obvious, how come A got to know that his
> number is 50? And no one answered it here... (probably, A cheated)
Assume the logicians know the others are logicians.
Tilford and Merriman have shown that (A, B, C) = (50, 20, 30) is the unique
solution.
To more easily see that it's a valid solution (uniqueness not shown by the
following argument) consider how A actually reasoned.
A saw B=20 and C=30, and thought:
A must be 10 or 50.
Assume A=10.
Then C would have seen A=10 and B=20, and would have thought:
"C must be 10 or 20.
If C=10 then B would have seen A=10 and C=10, and deduced B=20.
B said 'no', so C cannot be 10. Then C can only be 20."
C actually said 'no' so A deduces the assumption A=10 is false.
A then knows A=50.

Jens Kruse Andersen
> Assume the logicians know the others are logicians.
> Tilford and Merriman have shown that (A, B, C) = (50, 20, 30) is the unique
> solution.
> To more easily see that it's a valid solution (uniqueness not shown by the
> following argument) consider how A actually reasoned.
> A saw B=20 and C=30, and thought:
>
> A must be 10 or 50.
> Assume A=10.
> Then C would have seen A=10 and B=20, and would have thought:
> "C must be 10 or 20.
> If C=10 then B would have seen A=10 and C=10, and deduced B=20.
> B said 'no', so C cannot be 10. Then C can only be 20."
> C actually said 'no' so A deduces the assumption A=10 is false.
> A then knows A=50.
>
Best deduction. You really made it simple and straight.
Nishu

War doesn't determine who's right.
War determines who's left
Don't you mean "10 or 30"? (The argument still works though...)
Mark P wrote:
> Don't you mean "10 or 30"? (The argument still works though...)
Yes. I also noticed that typo.

Jens Kruse Andersen