How many wolves are there at the big round table?

Answer

Bye Marie

You can solve this by looking at the different possibilities and seeing which of them can match the data. I’ll try to outline it here by randomly starting at a point on the table. By W I mean a wolf, by G I mean a kid who speaks the truth, by g I mean a kid who is mistaken.

The table is round, so we should not consider special cases at the extremes.

It is best to start with a wolf (starting wolf, but that can be any wolf), because he always lies, while you are not sure about the goats.

If there is a wolf next to the ‘start wolf’, another wolf MUST be next to that wolf, otherwise the 2nd wolf would speak the truth and sit between a wolf and a goat. That 3rd wolf must therefore also sit next to a wolf, and so on. Then the table will look like this:

…WWWWWWW…

In other words, there are no goats at the table and there cannot be 2 who are mistaken. So next to a wolf can never sit another wolf.

So next to the (start) wolf there MUST be a goat, and on his other side too. After all, he must lie.

So we already have GWG (if there is no mistaken goat, but that will not be the case in most cases, since we have to get to 60 animals). So we will first continue with not-mistaken goats to see the pattern.

Next to the goat on the right next to the wolf, I have it G₁ called, a kid MUST G₂ sit down, otherwise G₁ not the truth. In addition to g₂ must a wolf sit, otherwise G₂ not the truth. In addition to that new wolf, a kid G₃ sit down, otherwise the wolf will speak the truth, next to G₃ must g₄ sit down, otherwise G₃ not the truth, and so on.

…GWG₁ G₂ WG₃ G₄ WGGWGGWG G…

You see that a set of 3 keeps coming back, WGG, where 1/3 is wolf and 2/3 is goat. In most cases this will have to be around the table, unless with the mistaken goats. So there are already 3.x animals around the table without the mistaken goats.

We now look at the different possibilities with the mistaken goats.

1) Suppose a mistaken goat comes after a wolf in our series, then another wolf must come after this goat, otherwise it would not be mistaken:

…GWGGWgWGGWGG…

So there is a set of 2 between the sets of 3.

2) Suppose a mistaken goat comes after a (not mistaken) goat, then another goat must come after this goat, otherwise it would not be mistaken.

a) Let’s first assume that this next one is a no-mistake kid. Then a wolf must come next, then another goat, and so on.

…GWGGWGGWGgGWGGWGG…

There is then a set of 4 between the sets of 3.

b) Suppose that the next mistake goat immediately follows the mistake goat, then it looks like this (There are only 2 mistake goats in total.):

…GWGGWGGWGggGWGGWGG…

There is then a set of 5 between the sets of 3.

We have a total of 60 animals around the table. Where are the mistaken goats?

If they are next to each other according to option 2b), then 5 places are occupied with WGggG. There are still 55 places left for sets of 3 WGG. 55 isn’t divisible by 3, so you can’t. The mistaken goats are not next to each other.

If there are 2 times a mistaken goat after a wolf, according to 1), then there are 2 sets of 2 (Wg). There are still 56 places left for sets of 3 WGG. 56 isn’t divisible by 3, so you can’t.

If there are 2 erroneous goats after a non-erroneous goat, according to 2a), then there are 2 sets of 4 (WGgG). There are still 52 places left for sets of 3 WGG. 52 isn’t divisible by 3, so you can’t.

If there is 1 time a mistaken goat after a wolf, according to 1), and 1 time a mistaken goat after a not mistaken goat, according to 2a), then there is a set of 2 (Wg) and a set of 4 (WGgG). There are still 54 places left for sets of 3 WGG. 54 IS divisible by 3 (18 sets of 3), and this is the only remaining possibility.

So we have 18 sets of 3 (WGG), in which there are 18 wolves.

Somewhere around the table Wg is sitting with 1 wolf.

Somewhere around the table is WGgG with 1 wolf.

That brings the total to 20 wolves. (They will eat well, with all those kids…)

I hope this is a bit clear.

Regards

Ineke De Coninck