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Mark with T or F all the below statements in such a way that they do not contradict with each other:

  1. At most $1$ statement is true
  2. At most $1$ statement is false

  3. At most $2$ statements are true

  4. At most $2$ statements are false

  5. At most $3$ statements are true

  6. At most $3$ statements are false

  7. At most $4$ statements are true

  8. At most $4$ statements are false

  9. At most $5$ statements are true

  10. At most $5$ statements are false

We have a total of $10$ statements. If we mark the $1^{\mathrm{st}}$ with T, it means all others must be false. But this contradicts with the 2nd. We can safely mark the last $2$ as True. But if we say, for example, at most $5$ statements are true, can it be also correct that "at most $3$ (or $2$ or $4$, etc.) statements are true"? At most $5$ statements are true means at least $5$ statements are false, which is OK in combination with the $10^{\mathrm{th}}$, right?

But if we say "at most $1$ statement is true", it means "at minimum $9$ statements are false". But if "at minimum $9$ statements are false", doesn't it mean also that "at minimum $7$ statements are false" for example?

How can we combine them all together?

I am completely stuck!

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  • $\begingroup$ Well, why not just try some cases and think about the implications? One trick that sometimes works: Assign values $T,F$ however you like, then evaluate each statement (reassigning according to whatever was true or false in your last iteration). If there is a consistent assignment it would be a fixed point for this operation. $\endgroup$
    – lulu
    Commented Feb 14, 2018 at 15:17
  • $\begingroup$ Note: somethings are clear. The first two statements can't both be true for example. More broadly, the number of true statements plus the number of false statements has to be $10$. That should help a lot. $\endgroup$
    – lulu
    Commented Feb 14, 2018 at 15:19
  • $\begingroup$ Dear Lulu, thank you for your comment. Obviously I have tried lots of combinations (I can't post them all here); I have been struggling with this problem for about a month now. $\endgroup$ Commented Feb 14, 2018 at 15:20
  • $\begingroup$ Well, have you tried my algorithm? Easy to automate...let a computer try to find a fixed point. $\endgroup$
    – lulu
    Commented Feb 14, 2018 at 15:22
  • $\begingroup$ I just tried it....found a fixed point almost instantly. $(0,0,0,0,1,0,1,0,1,0)$ where $1$ means True and $0$ means False. It's a good technique! You have to play with the seed (there are non-trivial cycles) and of course there might be more than one fixed point or some fixed points might be "repulsive". But still. $\endgroup$
    – lulu
    Commented Feb 14, 2018 at 15:25

1 Answer 1

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If we mark the 1st with T, it means all others must be false. But this contradicts with the 2nd. We can safely mark the last 2 as True.

I don't understand your reasoning here. Why are you sure you can safely mark 2 as "true?" You've just argued that 2 is false if 1 is true, so you need to be sure that 1 is false ...

(Now, 1 is in fact guaranteed to be false, but you haven't argued that yet.)

if we say, for example, at most 5 statements are true, can it be also correct that "at most 3 statements are true"?

Sure - maybe exactly 3 statements are true.


As a hint to get started, based on the two comments above:

Can you show that if the statement "at most $n$ statements are true" (resp. false) is true, then the statement "at most $k$ statements are true" (resp. false) is also true if $k>n$?

E.g. if 1 is true, then 3 has to be true as well, etc. This lets you rule out a bunch of statements right off the bat.

Another hint:

Try to solve a simpler case - say, what if you had only statements 1-4? What about 1-6?

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  • $\begingroup$ Thank you all guys, I think I am starting to understand! I still haven't found a "mathematical" way to prove Noah's statement above - I understand it by intuition only. $\endgroup$ Commented Feb 15, 2018 at 10:17

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