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I've recently heard a riddle, which looks quite simple, but I can't solve it.

A girl thinks of a number which is 1, 2, or 3, and a boy then gets to ask just one question about the number. The girl can only answer "Yes", "No", or "I don't know," and after the girl answers it, he knows what the number is. What is the question?

Note that the girl is professional in maths and knows EVERYTHING about these three numbers.


EDIT: The person who told me this just said the correct answer is:

"I'm also thinking of a number. It's either 1 or 2. Is my number less than yours?"

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    $\begingroup$ 'I'm also thinking of a number it's either 1 or 2, is my number bigger than yours?' does not work! It is not a solution $\endgroup$ – gota Oct 3 '13 at 13:04
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    $\begingroup$ I am thinking of a number greater than 1 and less than 3. Is my number greater than yours? -- That would work. $\endgroup$ – user1354557 Oct 3 '13 at 13:50
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    $\begingroup$ @ChibuezeOpata The trick is to pose the question with a new random value the girl doesn't know, thus resulting in the "I don't know". $\endgroup$ – SinisterMJ Oct 3 '13 at 16:19
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    $\begingroup$ Fix: I am thinking of a number that is either 1 or 2. Is my number greater than or equal to yours? $\endgroup$ – System.Cats.Lol Oct 3 '13 at 19:08
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    $\begingroup$ Why not simply I'm also thinking of a number. It's either 1.5 or 2.5. is my number bigger than yours? :) $\endgroup$ – peterwhy Oct 4 '13 at 5:36

39 Answers 39

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Assume the girl's number is X.

The boy asks:

Is half day before OP posted this question past October X?

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"Let your number be $n$, consider an equation $$x^n=n^2$$ If I randomly choose one of the real root(s) of $x$, is that root equal to your number $n$?"

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"Is $\Sigma$(TREE(number)) an odd number?"

NB: $\Sigma(\mathrm{TREE}(1)) = \Sigma(1) = 1$, $\Sigma(\mathrm{TREE}(2)) = \Sigma(3) = 6$, but in the context of "ordinary mathematics", the value of $\Sigma(n)$ is unprovable for any $n > 10\uparrow\uparrow10$, e.g. for $n = \mathrm{TREE}(3)$.

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  • $\begingroup$ While its precise value may be unprovable, you may have to work harder to demonstrate that its parity is not provable. $\endgroup$ – Ben Millwood Oct 4 '13 at 11:30
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Along the lines of "the open problem" method -

Define $$ f(n) = \pi^{n-1}\mathrm{e}^{\pi(n-1)}.$$

Where $n$ is the number chosen by the girl. The question is, is $f(n)$ irrational?

If $n=1$, $f(1) = 1$, so the answer is "No".

If $n=2$, $f(2) = \pi\mathrm{e}^{\pi}$, the answer is "Yes".

If $n=3$, $f(3) = \pi^2\mathrm{e}^{2\pi}$, so the answer is "I don't know".

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  • $\begingroup$ What if the girl knows the answer for the rationality of $\pi^2e^{2\pi}$? (And has just finished the paper, but not yet posted it online...) $\endgroup$ – Asaf Karagila Oct 8 '13 at 7:45
  • $\begingroup$ Sure, but I think the overwhelming consensus in this thread is that she has all current day knowledge but is not some omniprescient being. $\endgroup$ – Bennett Gardiner Oct 8 '13 at 12:31
  • $\begingroup$ Really? But the accepted solution (and the one suggested by the user) does not make the girl an omniscient being, knowing what we are thinking of? :-) $\endgroup$ – Asaf Karagila Oct 8 '13 at 12:38
  • $\begingroup$ Nope! Read them again, the girl needn't know what number you are thinking of. $\endgroup$ – Bennett Gardiner Oct 9 '13 at 5:37
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Let the boy ask the girl "Subtract 2 from the number your are thinking and then take the square root of the result.Is the result positive?". If the girl replies "yes" then the number is 3 because square root of 3-2=1 which is positive. If the girl replies "no" then the number is 2 because square root of 2-2=0 which is not positive.If the girl replies "I don't know " then the number is 1 because square root of 1-2= is imaginary number which cannot be said whether it is positive or negative.
OR

Let the boy ask the girl"Divide the number you have with the previous number.Ask if the result is a fraction?"If the girl replies "yes" the number is 3 because 3/2 is a fraction.If the girl replies "no" then the number is 2 because 2/2 is not a fraction.If the girl replies "I don't know" then the number is 1 because 1/0 is undefined.

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    $\begingroup$ Her answer would be "no" for 1, because the imaginary number isn't a positive number. It's also not a negative number. $\endgroup$ – Glen O Oct 3 '13 at 13:08
  • $\begingroup$ ask if the result is a natural number? $\endgroup$ – shrinath Oct 3 '13 at 13:32
  • $\begingroup$ That would be up to her definition of "natural number", since there's no defined convention. $\endgroup$ – Glen O Oct 3 '13 at 13:42
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    $\begingroup$ I think you're confusing 'I don't know' with 'neither'. $\endgroup$ – John Gowers Oct 3 '13 at 13:54
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If there is no generally agreed "$>$" operator for complex number field: "Is $e^{\left(i\frac{n-1}{2}\pi\right)}>0$?"

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Divide the other two numbers left to get a proper fraction. Ask her to subtract 1 from numerator and 2 from denominator. Ask her if the 'number' obtained is greater than 0.

If yes. She should have had 1 and thought of number 1. If she says she doesn't know. The number she thought of must be 3 If no. It should be 2

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Let $k$ be your your number. Is there a $3k$-colouring of the elements of $\mathbb{R}^2$ such that points having a unit distance have different colours?

Yes$\,\to k=3.\quad$No$\,\to k=1.\quad$I don't know$\,\to k=2.$

Have a look at the chromatic number of the plane.

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Call your number is $n$. Is $n=1$, or, is your reasoning assuming $n=3$ consistent?

If $n=1$ then trivially "yes".

If $n=2$, then $2=3$ is inconsistent, so "no".

If $n=3$, then by Godel she can't know the consistency of her own reasoning so "I don't know".

(Assuming her reasoning is consistent anyway...)

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protected by Community Oct 4 '13 at 5:43

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