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Liars, truth-tellers, mathematicians and physicists

Posted on: March 10th, 2014 by
6

There are N mathematicians and N physicists in a circular table. Some of them are liars and some are truth-tellers. The number of truth tellers who are mathematicians is equal to the number of truth teller a who are physicists. They are all asked, "Is the person next to you a mathematician or a physicist?" They all reply "physicist." Show that N must be even.

- via AMS math society


6 Responses to Liars, truth-tellers, mathematicians and physicists

  1. Lizard had this to say about that:

    They could be sitting in an alternating arrangement, with all mathematicians telling the truth and all physicists lying. This works for any N (odd or even).

    • Arpit had this to say about that:

      it is mentioned that the number of truth tellers that are mathematicians equals the number of truth tellers who are physicists. In your scenario, they are not.

  2. Sid Hollander had this to say about that:

    If n = 1 and Math (as always) tells the ruth and Physicist lies then they all will respond "Physicist".
    This appears to satisfy the conditions of the problem Yet n IS NOT even. How do YOU show that N is even?

  3. Mike Earnest had this to say about that:

    I assumed they were actually asked "Is the person TO YOUR RIGHT a mathematician or physicist?". Currently, the phrasing is ambiguous.

    The truth tellers are exactly those people who have a physicist on their right. Since there are N physicists, there are N people with physicists on their right, so there are N truth tellers. Since

    N = # Truth tellers = (# Math truth tellers) + (# Phys. truth tellers)

    and last two quantities are equal, we must have N is even.


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