On the island of Friends and Foes, every citizen is either a Friend (who always tells the truth) or a Foe (who always lies). Seven citizens are sitting in a circle. Each declares “I am sitting between two Foes”. How many Friends are there in the circle?

Clarification: You can assume that each person knows who is a Friend and who is a Foe.

This puzzle comes from the Senior Kangaroo 2022 contest.

  • $\begingroup$ Can you clarify "always lies." If a Foe sits between two Friends does he lie by negation saying "I am sitting between two foes" or can he say any lie other than the truth: "I am sitting between one/two foes"? $\endgroup$
    – akozi
    Commented May 16 at 16:05
  • $\begingroup$ @akozi The puzzle states that a Foe is someone who always lies. It does not specify how the Foe will lie. All you know is that a Foe will never speak a true statement. I hopes this helps, if not please feel free to ask again. $\endgroup$ Commented May 16 at 19:32

2 Answers 2


Solving generally, using notation (L)iars and (T)ruthers:

The chain is composed of multiple instances of only 2 possible elements:

LT and LLT. Reasoning:

You can't have LLL, else the Liar's statement would be truth

You can't have TT, else the Truther's statement would be false

These elements both have one truther in them, and one is of length 2 and another is of length 3.

Thus, the question becomes: "How many addends of either 2 or 3 can an addition have to sum to N?"

The only answer for N=7 is exactly three: 2+2+3

  • $\begingroup$ Cool reasoning! $\endgroup$ Commented May 16 at 19:42
  • $\begingroup$ While the answer is correct, I dont think this answer is 100% complete. You can have for example TL instead of LT. And those 2 can potentially 'merge' into ....TLTLTL...., about which you have nothing said. And so, it could potentially be 4 truthers. $\endgroup$
    – Lezzup
    Commented May 19 at 17:20
  • $\begingroup$ @Lezzup I did not include the fact that if you start with "T" you should shift. TLTLTLT is equivalent to LTLTLTT which has TT which is illegal $\endgroup$ Commented May 20 at 10:16
  • $\begingroup$ I know why 3 is maximum, I know that with TLTLT... you end up with TT. I am just pointing out it should be included in the reasoning. $\endgroup$
    – Lezzup
    Commented May 20 at 10:54

The answer is

Three Friends


Three is the maximum number of Friends. Each Friend must be between two Foes. There is no way to place 4 people in a circle of seven where none of them are adjacent. You can place 3 people without adjacencies, so the maximum number of Friends is three. Three is also the minimum number of Friends. Each Foe must be adjacent to at least 1 Friend, and if two friends are placed as far apart as possible, there will be two Foes between them on one side, and three on the other. The middle of the three Foes is not adjacent to any Friends. So there must be at minimum three Friends. Since there must be at minimum three Friends, and at maximum three Friends, there are three Friends. Maybe d'Artagnan should see if they're recruiting?


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