Somewhere on the Hyperbolic Plane, you encounter ten mathematicians. You know that some of them always tell the truth while the rest always lie. Each of them knows which of the others tell the truth and which do not.

Let $g$ be the number of mathematicians that tell the truth. Let $h$ be the sum of the indices of the mathematicians that tell the truth (e.g. if mathematicians 4 and 5 tell the truth, $h=4+5=9$).

You ask the mathematicians for the values of $g$ and $h$ and receive the following answers:

Mathematician   1: $\quad g = 6 \quad \lor \quad h = 12$

Mathematician   2: $\quad g = 3 \quad \lor \quad h = 12$

Mathematician   3: $\quad g = 4 \quad \lor \quad h = 11$

Mathematician   4: $\quad g = 4 \quad \lor \quad h = 25$

Mathematician   5: $\quad g = 2 \quad \lor \quad h = 17$

Mathematician   6: $\quad g = 3 \quad \lor \quad h = 24$

Mathematician   7: $\quad g = 5 \quad \lor \quad h = 23$

Mathematician   8: $\quad g = 5 \quad \lor \quad h = 11$

Mathematician   9: $\quad g = 6 \quad \lor \quad h = 23$

Mathematician 10: $\quad g = 5 \quad \lor \quad h = 16$

Q: Which of the mathematicians are telling the truth?

Note 1: $\lor$ is the boolean "or" operator, so only one of the statements needs to be correct in order for a mathematician to be telling the truth.

Note 2: I constructed this puzzle using a search program that enumerates all possibilities to ensure correctness. It would therefore be nice to get solutions that do not rely on computer programs themselves.

  • $\begingroup$ By note 1, can i say that the lying mathematician also have just one lie out of both? $\endgroup$
    – manshu
    Jan 10, 2016 at 17:00
  • 1
    $\begingroup$ @manshu: No. $a \lor b$ is true iff at least one of $a,b$ is true. If just one of $a,b$ is a lie, $a \lor b$ is still true. $\endgroup$
    – user18360
    Jan 10, 2016 at 17:14
  • 1
    $\begingroup$ I don't understand this at all - the mathematicians either always lie or always tell the truth, and yet you have constructed this so that a mathematician can not only do both, but be considered to be truthful in so doing? $\endgroup$ Jan 11, 2016 at 14:57
  • 2
    $\begingroup$ @question_asker: It is not immaterial. "The sky is blue or green" is a true statement, and someone who always tells the truth could easily say it. $\endgroup$ Jan 11, 2016 at 15:49
  • 2
    $\begingroup$ @question_asker You calling mathematical logic something from the "Obnoxious Nerd perspective" may or may not (a tautologically true statement!!) give you street cred, but mathematicians ain't gonna buy it. BTW I'd love to play some gambling games with you. :) ) $\endgroup$
    – Paul Evans
    Jan 11, 2016 at 21:39

3 Answers 3


By case distinction on the value $g$:

  • If $g=1$, then the unique truth teller Mx must claim that $h=x$; contradiction.
  • If $g=2$, then M5 tells the truth ($g=2$). The other truth teller Mx must claim that $h=x+5$; contradiction.
  • If $g=3$, then M2 and M6 tell the truth ($g=3$). The other truth teller Mx must claim that $h=x+2+6=x+8$; contradiction.
  • If $g=4$, then M3 and M4 tell the truth ($g=4$). The other two truth tellers Mx and My must claim that $h=x+y+7$; in particular, they both must claim the same value of $h$. The pairs (M1 and M2) and (M3 and M8) do not work. Only (M7 and M9) works.
  • If $g=5$, then M7, M8, M10 tell the truth. The other two truth tellers Mx and My must claim that $h=x+y+25$; this value $h$ is too large.
  • If $g\ge6$, then at least four guys must state the correct value of $h$; contradiction.

To summarize:

M3, M4, M7 and M9 tell the truth with $g=4$ and $h=23$.

Furthermore, there is a second solution:

None of them tells the truth with $g=0$ and $h=0$. (For this, you must accept that the empty sum takes value $0$.)

  • $\begingroup$ Second solution is fascinating. $\endgroup$ Jan 10, 2016 at 17:50
  • 4
    $\begingroup$ @AycanYaşıt Second solution is invalid, the author already specified that some of them always tell the truth. $\endgroup$
    – March Ho
    Jan 10, 2016 at 23:21
  • $\begingroup$ @Ninety-Three The answer is only valid if none of them tell the truth, not all of them. $\endgroup$
    – March Ho
    Jan 11, 2016 at 3:03
  • $\begingroup$ Just as a note: When doing this, I found it to be simpler to reduce everything using the obvious information. E.g. If 8 is correct then there is $g*=6-1=5$ other people telling the truth or the sum of the rest is $h*=11-8=3$ (Basically this directly eliminates $x$ from your reasoning) $\endgroup$
    – Bort
    Jan 11, 2016 at 8:47

By iterating through the possible values of g, then filling the remaining missing mathematicians who use the correct value h, I was able to determine another solution:

Mathematicians 2 and 6 claimed the value g = 3, and by iterating through the remaining mathematicians, Mathematician 3 claimed the value h = 11. 3 mathematicians, and sum of indices = 11.


Upon further consideration, my answer is actually invalid:

If this solution is valid, then Mathematician 8 is also telling the truth (e.g. h = 11) which cannot be, so this is a contradiction and is not a valid solution.

  • 4
    $\begingroup$ Looks like I tricked 2 people $\endgroup$ Jan 10, 2016 at 20:57

The truth tellers are:

M3, M4, M7 and M9


Can't be 2 truth tellers because that would mean M5 and someone else. All the others don't have their index plus 5 equal to their stated h, so no go.
Likewise there can't be g = 3 because we have M2 and M6. But there's no single other mathematician with h = index + 2 + 6, so no go.
Can't be g = 5 as 7 + 8 + 9 = 24. Minimum other two h's must be greater than be 27. No h > 27, so no go.
Can't be g = 6 as then we'd have M1 and M9. We'd then need 4 identical h's which isn't so, so no go.
This leaves g = 4 which gives us M3 and M4 with g = 4 and M7 and M9 with h = 3 + 4 + 7 + 9

Sorry output's so ugly. Can't get spoilers to newline properly. Thank you @The Dark Truth

Just started looking at other answers and from @Gamow saw I missed g = 1 and g >= 7

Apologies to @Gamow for stealing M[0-9]+ notation.

  • $\begingroup$ I made the proper line breaks for your spoilers. Now you should work on correcting your spelling. $\endgroup$ Jan 11, 2016 at 14:56
  • $\begingroup$ @TheDarkTruth Thanks, how'd you do it? $\endgroup$
    – Paul Evans
    Jan 11, 2016 at 14:59
  • $\begingroup$ Secret :P (Put two spaces at the end of each line) $\endgroup$ Jan 11, 2016 at 15:06
  • 1
    $\begingroup$ Ah! Thank you master, won't tell a soul ;) $\endgroup$
    – Paul Evans
    Jan 11, 2016 at 15:08

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