Three logicians...

15 views
Skip to first unread message

lightofreason

unread,
Apr 22, 2006, 7:27:06 PM4/22/06
to
Does anybody know the solution to this puzzle?...

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?

Stephen Merriman

unread,
Apr 22, 2006, 9:01:21 PM4/22/06
to

spoiler
.
.
.
.
.
.
.
.
.
-
.
.
.
.
.
.
.
.
.
-
.
.
.
.
.
.
.
.
.
-
.
.
.
.
.
.
.
.
.
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, |y-z| are possible: ie
y!=z.
The second person saying no must mean both of x+z, |x-z| are possible:
ie x!=z, and x!=2z.
The third person saying no must mean both of x+y, |x-y| are possible:
So x!=y, and if x>y then:
x-y != x or x/2, so x!=2y
if x<y then:
y-x != 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), |y-z| must be impossible, and x = y+z.
If y>z, we must have y-z = 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 z-y = 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 = x-y = 30.

Stephen

Arthur J. O'Dwyer

unread,
Apr 22, 2006, 9:04:39 PM4/22/06
to


PARTIAL

S
P
O
I
L
E
R


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 go-round,
so I would guess that the other two numbers were 24 and 26. Not sure
how to prove it, though.

-Arthur

Pitbull

unread,
Apr 22, 2006, 9:12:30 PM4/22/06
to
"lightofreason" <lighto...@gmail.com> wrote in
news:1145748426....@i40g2000cwc.googlegroups.com:


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

Mark J. Tilford

unread,
Apr 22, 2006, 10:15:50 PM4/22/06
to


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

Nishu

unread,
Apr 23, 2006, 11:28:27 PM4/23/06
to

Mark J. Tilford answered:

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?

Stephen Merriman

unread,
Apr 24, 2006, 1:02:06 AM4/24/06
to
Nishu wrote:
> 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

Jens Kruse Andersen

unread,
Apr 24, 2006, 9:44:41 AM4/24/06
to
> > lightofreason <lighto...@gmail.com> wrote:
> > >
> > > 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?

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


Nishu

unread,
Apr 24, 2006, 11:44:59 PM4/24/06
to
Jens Kruse Andersen answered:

> 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

Mark P

unread,
Apr 25, 2006, 2:37:15 PM4/25/06
to
Jens Kruse Andersen wrote:
>>> lightofreason <lighto...@gmail.com> wrote:
>>>> 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?
>
> 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.

Don't you mean "10 or 30"? (The argument still works though...)

Jens Kruse Andersen

unread,
Apr 25, 2006, 6:51:25 PM4/25/06
to
> > Then C would have seen A=10 and B=20, and would have thought:
> > "C must be 10 or 20.

Mark P wrote:
> Don't you mean "10 or 30"? (The argument still works though...)

Yes. I also noticed that typo.

--
Jens Kruse Andersen


Reply all
Reply to author
Forward
0 new messages