SOLUTIONS

Puzzles

  1. The 7 Monks
  2. How old are my children ?
  3. Kitty Race

Paradoxes

  1. The hangman

SOLUTION FOR THE 7 MONKS

Consider the following solution:

Suppose there was only one monk there. Since he knows that at least one person has to be sick, he knows that it has to be him and he leaves the monastry.

The case with two monks.

Suppose only one monk is sick. On the first evening he looks at the other monk and he does not see a black spot on his forehead, so the other one is not sick and since there is at least one sick monk, he concludes that it has to be him and leaves the monastry. In the case that they are both sick they will both see a black spot on the other one's face. The next day they see that the other person is still there. Then they know that they are both sick because otherwise the other one would have left the monastry on the first day. So, on the second day they both leave the monastry.

...?...The case with N monks.

Reasoning by induction we know that in the case of seven sick monks, all seven will leave the monastry on the seventh day.

What's wrong with this solution ?

This solution seems to be perfectly reasonable and for the last 10 years I tought that this solution was correct. But it is not (or maybe not)! Consider the case where all the monks just ignore all other monks for the first three days before they start reasoning like above. The result is that it will take three days longer until they know they are sick, but this way of thinking of the monks will also produce the wanted result. But now we run into troubles. Suppose that the first monk uses the first method and the other monks use the technique with a delay of three days. In this case the first monk will make the wrong assumptions and he will fail to know correctly if he is ill or not. The problem is that we make an extra assumption in our solution, that is:

"Every monk uses the same strategie to solve the problem"

For our solution to be correct we have to prove that this solution is the only possible one which clearly is not the case because I just gave a counter example.

What I am wondering about now is:

If I add the condition that the monks have to use the fastest strategie possible, is this problem then solvable or not? In other words, can you prove that the given strategie is the fastest one there is?

(Since I do not know the answer I would be glad to get suggestions about it. You can mail me right here)

Note: I got a solution from Matthias Kirsch that delivers an immediate solution:

>
> I have thought of a solution which gives instant results:
>
> n sick monks see (n-1) monks with black spots; the monks which are not
> sick see n black spots; now both groups have to consider that they are
> sick as well; so the goup which is well has to think that there are
> either n or n+1 ill monks; they also have to take into account their sick
> fellow monks and know that their range is either n or n-1; the same holds
> true for the sick monks.
>
> Now what if they all act unisono and conclude, completely logical that
> all the monks which see n-1 spots have to kill themselves instantly. If
> there are 7 sick monks all the ones which see six spots will have to
> commit suicide -> seven monks will die. I do not understand why it
> works, but it does.
>
> what do you think. Sincerely Matthias Kirsch

This solution is fully correct if you assume that n is known. This is not needed for the other solution, the only thing that is needed in the previous solution is the knowledge that at least one monck is sick. A quick look to my page revealed that I did not make this point clear on my page, I even never considered it necessary to do this. But confronted with this solution it is clear that it is an important point, so I added it to my page. The same solution was also found by Aidan McKenna.

SOLUTION FOR HOW OLD ARE MY CHILDREN ?

Fact 1 : The product of their ages is 36

Resulting posibilities.

( 1, 1,36) ( 1, 2,18) ( 1, 3,12) ( 1, 4, 9) ( 1, 6, 6)
( 1, 9, 4) ( 1,12, 3) ( 1,18, 2) ( 1,36, 1) ( 2, 1,18)
( 2, 2, 9) ( 2, 3, 6) ( 2, 6, 3) ( 2, 9, 2) ( 2,18, 1)
( 3, 1,12) ( 3, 2, 6) ( 3, 3, 4) ( 3, 4, 3) ( 3, 6, 2)
( 3,12, 1) ( 4, 1, 9) ( 4, 3, 3) ( 4, 9, 1) ( 6, 1, 6)
( 6, 2, 3) ( 6, 3, 2) ( 6, 6, 1) ( 9, 1, 4) ( 9, 2, 2)
( 9, 4, 1) (12, 1, 3) (12, 3, 1) (18, 1, 2) (18, 2, 1)
(36, 1, 1)

Removing all doubles (for example 1,1,36 and 36,1,1) and adding up gives :

( 1, 1,36) = 38
( 1, 2,18) = 21
( 1, 3,12) = 16
( 1, 4, 9) = 14
( 1, 6, 6) = 13
( 2, 2, 9) = 13
( 2, 3, 6) = 11
( 3, 3, 4) = 10

Fact 2: Knowing the housenumber and knowing that the sum of their ages equals the housenumber, the second friend still couldn't solve the problem.

So the housenumber has to be 13, because only this housenumber leaves two solutions left.

( 1, 6, 6) = 13
( 2, 2, 9) = 13

Fact 3: The eldest one was with his grandmother.

So 1, 6, 6 is excluded because this combination hasn't got an eldest one.

Solution

The ages of the children are 2, 2 and 9 and the housenumber was 13

Kitty Race

This was an easy one, at least when you start it the right way ! The small kittens have to run 50 meters. They run at a speed of 1 m/s, so it takes 50 seconds for the kittens to reach the middle. In the mean time the mother is running at a speed of 10 m/s resulting in 50 s x 10 m/s = 500 meter ! (If you solved the problem this way you are maybe thinking: "So what's the big deal ?". Well, nothing, but most off the people start calculating sums of the distance between the kittens and the mother making this a very difficult problem to solve)

PARADOX DISCUSIONS

 

THE HANGMAN

 The problem with this paradox is the artificial division of time. If the judge said, it will be a surprise, he did not meant it to be a surprise in the very last second before he died, because everybody knows that such a statement is not correct (the rope is already around his neck and they just pulled the lever). So in the continous case, it is clear that the judge made a mistake by claiming Sam's dead as a surprise. We think that the only thing what happens here is that we force people to think in days as the smallest amount of time. If we do that then there is no reason why we can not hang Sam on Sunday because it has been a surprise UNTIL saturday (compare with the very last second in the continuous case), so the judge was right when Sam is killed on Sunday, that Sam was killed by surprise, because Sam couldn't know which day the judge had chosen until saturday, i.e. the smallest time unit before he is hang.