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In the land of Truthlandia, each person is either a truth teller who always tells the truth, or a liar who always tells lies. All 33 people who gathered for a meal in Truthlandia at a round table, said: "The next 10 people on my right are all liars." How many liars were actually in attendance?

I have looked around quite a lot, but the closest I could find was this. If anybody can answer that would be greatly appreciated. Thanks for any answers!

Parcly Taxel
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Jmaxmanblue
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    Nice puzzle question. If you haven’t already, you might be interested in visiting the sister-site here on SE: puzzlingSE A lot of similar - and a lot of very different - puzzles like this plus a hole community loving these kind of things and both inventing and solving them. – BmyGuest Dec 01 '17 at 06:29
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    One: the storyteller. There's no such land. – Ooker Dec 01 '17 at 11:58
  • This is certainly not math. – Martin Argerami Dec 01 '17 at 21:53
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    @BmyGuest Thanks for the tip. Puzzling SE seems like the place to post it, but it's too late now. If I have any further questions like this then I will definitely go there! – Jmaxmanblue Dec 01 '17 at 22:26

4 Answers4

37

Someone must be telling the truth, or else 33 liars would all be claiming that the next ten people are liars—and they would all be correct.

If Person A tells the truth, then A is correct that the next ten people are all liars:

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In particular, we know that Person B, the person on Person A's right side, is a liar. Person B claims that the next ten people are all liars:

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Person B must be wrong about that claim, because Person B is a liar. The first nine people are liars because A said they were. This implies that the last person isn't a liar --- the tenth person after B is honest:

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and now the reasoning repeats again with a new truth teller. In the end, you're left with three equally-spaced truth-tellers amidst 30 liars.

In general, you can have a circle of $n$ people claiming that the next $k$ people are all liars. If $n$ is a multiple of $k+1$, then the situation is solvable: there's one truth teller followed by $k$ liars, followed by a truth teller, etc. If $n$ isn't a multiple of $k+1$, then you can't solve the problem in Truthlandia. (Compare two versus three people sitting around a table all claiming that the next person is a liar. The situation is solvable with two people, but not with three.)

user326210
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    Nice answer! But in theory, you would still need to prove that there exists at least one truth teller, isn't it? – Surb Dec 01 '17 at 11:19
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    @Surb If everyone was a liar, we'd have a table full of liars all telling the truth, which they don't do. So we're guaranteed to have at least one truth-teller. – doppelgreener Dec 01 '17 at 12:11
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A truth-teller, by the given statement, forces the nature of the ten people to the right – a block of eleven. However, there cannot be a block of eleven liars, because the leftmost "liar" would actually be telling the truth. Thus there are three equally-spaced truth tellers and 30 liars.

Parcly Taxel
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Notice that not everyone can be a liar because in that case the statement about the people to their right is actually true. This guy will give us everything we need in domino-like fashion.

Number the people $P_1,\cdots P_{33}$ say sitting in clockwise order and say $P_1$ is a truth teller. This forces $P_2,\cdots, P_{11}$ to all be liars. Now $P_2$ is a liar so this forces that there has to be a truth teller in the next ten set: $\{P_3,\cdots ,P_{12}\}$. Together, we see that $P_{12}$ is a truth teller so $P_{13},\cdots ,P_{22} $ are all liars.

By a similar argument as above, $P_{23}$ is a truth teller and so this finishes the argument since now $P_{24},\cdots ,P_{33}$ are all liars.

So exactly three truth tellers and $30$ liars.

Arkady
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2

Here is alternate view point.

The table was against the wall an nobody was on the right side.

They all lied by claiming there were 10 people were on the right. If they were all liers they would have to lied about that also. If your lying you might as might as well blame non-existent people because they can't complain about it.

In fact, The number 10 is a lie because they have to ALWAYS lie. Also right has to be a lie so left side or non-existant for the same reason.

cybernard
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  • at a round table. – StainlessSteelRat Dec 01 '17 at 20:49
  • @StainlessSteelRat Maybe half the round table is unused. The liars lie about everything so they would have to lie about the number of liers it part of lying. – cybernard Dec 01 '17 at 20:56
  • @StainlessSteelRat In fact they ALWAYS lie implying the liers on the left side because they said the liers were on the right. – cybernard Dec 01 '17 at 21:06
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    Now, you are just putting lies in their mouths. These are only solved if we stick to basic rules. – StainlessSteelRat Dec 01 '17 at 21:35
  • @StainlessSteelRat In real life nobody likes a lier so the true tellers wouldn't socialize with the liers. Mine solves 33 liers, and the rest of the truthful people went to an honest pub. You gonna pick up the tab? Yes... , whoops I lied. You a designated driver, whoops? Yes ...I lied. You gonna tell my wife/girlfriend I was at the bar all night? No... whoops I lied. – cybernard Dec 01 '17 at 22:24
  • Math is not about lies. 8) – StainlessSteelRat Dec 01 '17 at 22:28
  • @StainlessSteelRat No math is all about the facts and/or true. And the true here is we have 33 liars. The problem evens says they ALWAYS lie. – cybernard Dec 01 '17 at 22:30