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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?"

Mike Pierce
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Gintas K
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  • So basically, each of the answer links to the number, but what should he ask, that she wouldn't know about one of the numbers? – Gintas K Oct 03 '13 at 09:24
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    '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 – gota Oct 03 '13 at 13:04
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    Your Note does not agree with your comment. If the girl "knows EVERYTHING about those three numbers", then surely there is no possible solution to "what should he ask, that she wouldn't know about one of those numbers?". – r.e.s. Oct 03 '13 at 13:30
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    I am thinking of a number greater than 1 and less than 3. Is my number greater than yours? -- That would work. – user1354557 Oct 03 '13 at 13:50
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    Normally, if someone says 'I don't know', when they actually know, it's usually because they are withholding the truth (answer). Why assume otherwise here? – Chibueze Opata Oct 03 '13 at 15:32
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    @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". – SinisterMJ Oct 03 '13 at 16:19
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    Would someone mind turning the answer at the end of OP's post into a spoiler? – Cameron Williams Oct 03 '13 at 16:19
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    Fix: I am thinking of a number that is either 1 or 2. Is my number greater than or equal to yours? – System.Cats.Lol Oct 03 '13 at 19:08
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    Fix: What is your number; if it is one then answer YES, if it is two then answer NO, if it is three then answer I don't know? – kobaltz Oct 03 '13 at 20:54
  • "I'm also thinking of a number. It's either 1 or 2. is my number bigger than yours?" - Won't work. The answer would be "NO" if the girl thinks 1 or even if she thinks 2. Its kind of programmer's logic. – TheRookierLearner Oct 04 '13 at 02:30
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    @TheRookierLearner, if the girl's number is 1, then her answer would be "I don't know" (1 is not greater than 1, 2 is greater than 1). If the girl's number is 2, then her answer would be "No" (1 is not greater than 2, 2 is not greater than 2). If the girl's number is 3, then her answer would be "No". So the OP's proposed answer is incorrect. – JRN Oct 04 '13 at 03:19
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    Why not simply I'm also thinking of a number. It's either 1.5 or 2.5. is my number bigger than yours? :) – peterwhy Oct 04 '13 at 05:36
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    Please say yes if your number is 1, no if your number is 2, i dont know if your number is 3. – dust05 Oct 05 '13 at 04:18
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    The OP has fixed their answer and it is correct. Another one that works in the same way is: I've written down either 2 or 3, is my number greater than yours? If the girl has 1 then 2 and 3 are both greater, so 'Yes'. If she has 2 then 2 is not greater but 3 is, so 'Don't know'. If she has 3 then neither 2 nor 3 are greater than 3, so 'No'. @user1354557, choosing from the uncountable continuum between 1 and 3 is just overkill. And while I'm at it, there are too many silly answers. – Geoff Pointer Oct 10 '13 at 05:19
  • Why not simply, if your number is 1, answer yes, if it 2, answer no and if it is 3 answer I don't know? – Sawarnik Nov 09 '13 at 17:56
  • Let's say "Yes" denotes 1, "No" denotes 2, "I don't know" denotes 3, what's your number? – chaohuang Mar 26 '14 at 22:23
  • Ask her if the number multiplied by 11 can be written as the sum of three cubes of integers. – Marijn Nov 30 '15 at 00:11
  • Personally I liked: Lady or the Tiger? And Other Logic Puzzles Including a Mathematical Novel That Features Godel's Great Discovery--by Raymond M. Smullyan – rrogers Dec 09 '15 at 13:34
  • Godel might disagree that this girl knows everything about these numbers. I'd suggest restricting her computational ability to something like primitive recursion. – DanielV Feb 22 '19 at 01:02

39 Answers39

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"I am thinking of a number which is either 0 or 1. Is the sum of our numbers greater than 2?"

Ben Millwood
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    +1. Your answer appeared while I was composing my answer. This kind of approach beats appealing to currently-unsolved aspects of number theory, because the "I don't know" response is given when the answer is actually unknowable not merely unknown for now. There's no expiration date on this solution. – Blue Oct 03 '13 at 10:45
  • I don't get, why do you have to think of a number yet your one number is not fixed? For me, the most puzzling thing about this question is, if the person is going to tell you the truth, why not just ask them the number. If they want it mathematically, why use a cheap solution like this? – Chibueze Opata Oct 03 '13 at 15:30
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    You have to offer two options to allow each of her three choices to correspond to one of the numbers. If her number is 3 she'll say Yes because adding 3 to 0 or 1 is always greater than 2. If her number is 2 she won't know, because if yours is 0 it won't be but if it's 1 it would be. If her number is 1 then adding either 1 or 0 won't make it greater than 2. – deed02392 Oct 03 '13 at 15:44
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    @ChibuezeOpata You're missing the point. The person can only answer "Yes" "no" or "I don't know" and must do so honestly. You can't ask "what's the number" because that's an invalid question within the constructs of the riddle. – xdumaine Oct 03 '13 at 16:13
  • @xdumaine: I get it, but don't you prefer Peter Olsen's answer ? – Chibueze Opata Oct 03 '13 at 16:35
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    @ChibuezeOpata No, I don't prefer it. This answer is in the spirit of the question while that one uses a trick of language to contrive an answer. – xdumaine Oct 03 '13 at 16:40
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    Although to be technical, it violates the constraints of the riddle. The riddle says you may ask one question, it does not state that you can ask a question dependent on a statement. – JohnP Oct 03 '13 at 17:09
  • What if the girl doesn't know how to add, and says "I don't know?" – sksallaj Oct 03 '13 at 18:25
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    @JohnP couldn't it be easily altered? I.e. "Is the sum of your number and a number uniformly chosen from ${0,1}$ greater than 2?" – Robert Mastragostino Oct 03 '13 at 22:54
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    @RobertMastragostino - Yes, it could. As written though... (I'm much better at English than teh maths) – JohnP Oct 03 '13 at 22:55
  • Everyone starts with "I am also thinking of two numbers..". Won't the girl require to know the boy's number before answering? – TheRookierLearner Oct 04 '13 at 02:33
  • @TheRookierLearner No, because it doesn't matter. – Georgian Oct 04 '13 at 21:31
  • @GGrec - Could you elaborate? Thanks! – TheRookierLearner Oct 04 '13 at 23:58
  • @TheRookierLearner If she says YES, then the answer is 3 (0+3=3; 1+3=4). If she says NO, then the answer is 1 (0+1=1; 1+1=2). If she says I DON'T KNOW, then the answer is 2 (0+2=2; 1+2=3). – Georgian Oct 05 '13 at 09:33
  • Simple? Maybe, but who says that simpler = better? As for me, I dont see how is this better than many other answers here. – Adam Oct 07 '13 at 15:16
  • This answer is not really any different in principle to the one offered by the OP. The general type is "I'm thinking of one of two numbers, how does yours relate to mine?" Another that works is "I'm thinking of 2 or 3, is my number greater than yours?". As JohnP infers above, what exactly does one question mean? Personally, I'm still waiting to see a really stunning solution to this riddle - if one exists ... – Geoff Pointer Oct 10 '13 at 23:21
  • Since the girl knows EVERYTHING about number 1, she knows if the boy is thinking to that number... – Emanuele Paolini Apr 08 '15 at 09:05
139

"I'm also thinking of one of these numbers. Is your number, raised to my number, bigger than $2$?"

Let $n$ be the girl's number (unknown to me), and let $m$ be my number (unknown to her).

  • $n = 1 \implies $ NO: $1^m = 1 \not > 2$ for all $m \in \{1,2,3\}$.
  • $n = 3 \implies $ YES: $3^m \geq 3 > 2$ for all $m \in \{1,2,3\}$.
  • $n = 2 \implies $ I DON'T KNOW: Whether $2^m > 2$ depends on $m$.
Blue
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First of all I ruled out indirect ways of using reference to either of the numbers 1, 2 ,3 to frame a question, as I thought it's implicit in the question that it should challenge your thinking, not your cleverness. If she answers I don't know, compared to yes or no, it is more likely that she is confused between two numbers, ruling out one possibility. If the boy thinks of a number, the most common way to link it up to 1, 2, or 3 will be by divisibility. Also, I thought out of 1, 2, and 3, if I can rule out one number by the way I frame the question, I will be left with two options. The most common way of describing a number is whether it's odd or even.

So how about asking: "I am thinking of an odd number. Is it perfectly divisible by your number?"

If she says no, clearly the number is 2. If she says yes the number is 1 because only with 1 can you be sure that any number is divisible by 1. If she says I don't know the number is 3, because an odd number can or cannot be divisible by 3.

Mike Pierce
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ND_27
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Pick any bijective mapping from $\{1,2,3\}$ to $\{\text{Yes},\text{No},\text{I don't know}\}$ and then it is easy to contrive a question.

Here's an example question using this method.

Let $f =\{(1,\text{Yes}),(2,\text{No}),(3,\text{I don't know})\}$ and let $x$ be the number that you are thinking of. What is $f(x)$?

This can be generalized to any set of possible numbers and set of possible responses with the same cardinality.

Peter Olson
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    This seems to violate the purpose of a yes-no(-I-don't-know) question. A yes-no question is characterized by the question word, not by the answer. So you shouldn't use question words "what, when, where, why, who, or how". One way to know if your answer works is to ensure that it stilll works if she decides to answer one of "affirmative", "negative", "absolutely", "of course", "I can't tell" etc. – dspyz Oct 03 '13 at 18:45
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    @dspyz I understand what you're saying, and I will admit that this is likely not how the question was designed to be interpreted. That said, the question does state, The girl can only answer "Yes", "No" or "I don't know", and does not make any restrictions about what those responses have to mean. – Peter Olson Oct 03 '13 at 18:51
  • Good point, it doesn't actually say anywhere that the question has to be a "yes-no" question. I should read it more carefully next time. – dspyz Oct 03 '13 at 19:03
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    I like this answer, but it's cheating not to ask a yes-or-no question and to employ the knowledge part of the puzzle. – Caleb Stanford Nov 10 '16 at 23:42
  • Ultimately the riddle is asking you to SELECT such a bijection $f$ that induces a well-formed "yes/no" question. – Ehsaan Oct 20 '21 at 17:32
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Let the boy ask the girl:

"Divide the number you have with the previous number. Is the result a fraction?"

If the girl replies:

  • "Yes" then the number is 3 because 3/2 is a fraction.
  • "No" then the number is 2 because 2/1 is not a fraction.
  • "I don't know" then the number is 1 because 1/0 is undefined.
Nick Orlando
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shrinath
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    I love most of these answers (great ways to think about riddles), but this one is especially clever because of the undefined factor (1/0). Great answer. – Question3CPO Oct 03 '13 at 14:12
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    I disagree. The correct answer in the 1 case is "I can't do that." – Ben Millwood Oct 03 '13 at 17:59
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    Integers are fractions too. The denominator is 1 in lowest terms, but it's still a fraction. He should ask "Is the number an integer?" Also, the question of how to handle 1/0 depends on what space you're working in. True, there's no real number 1/0, but how do you know she's not thinking of the Reimann Sphere (in which case "no", infinity is not an integer).

    And if you say "Is the resulting real number a fraction?" Then the correct answer is "your question makes a false assumption"

     – dspyz Oct 03 '13 at 18:53
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    "Is 1/0 a fraction?" The answer is certainly "no", not "I don't know". – 2'5 9'2 Oct 04 '13 at 06:36
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    $\frac{1}{0}$ is undefined, but we know that it's not a frog, an icecream, or a differentiable function. Undefined is not the same as SQL NULL. – jwg Oct 04 '13 at 08:34
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    Can you prove that 1/0 is not a frog? – Richard Venable Oct 04 '13 at 19:09
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    Undefined is a very precise mathematical term. It does not mean: "We don't know". We can say that 1/0 will never be defined because in terms of our definition of multiplication of numbers, no sense can be made of dividing by zero. The real numbers, without zero, are a multiplicative group. We don't do this because we haven't yet figured out how to find a multiplicative inverse of zero. It's not an unsolved problem. In limit theory, when we get 1/0, the answer is: "Does not exist". "I don't know" is not the correct answer, so this question doesn't work. – Geoff Pointer Oct 10 '13 at 23:40
  • @RichardVenable 1/0 is not defined and does not exist. A frog is defined and does exist. Piddle about with nit picking philosophical arguments if you like but any reasonable mathematician can tell the difference between 1/0 and a frog. – Geoff Pointer Oct 10 '13 at 23:42
  • I don't agree that we know that 1/0 is not a frog. If I define 1/0 as being a certain individual frog, then my definition doesn't contradict any mathematical theorem. It's not an invalid definition; it's merely a silly and nonsensical one. And doesn't "undefined" just mean "not having a valid, agreed-upon definition"? – Tanner Swett Oct 18 '13 at 04:41
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"Among all prime numbers except 3, is there a positive and finite number of couples whose difference is the number you are thinking of?"

$N = 1 \implies$ no, since there are none ( $\{2, 3\}$ is disallowed).

$N = 2 \implies$ I don't know, at least until the Twin Primes conjecture is proved or disproved.

$N = 3 \implies$ yes, since there is only $\{2, 5\}$.

John Gowers
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mau
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    While we're doing the conjecture thing, we could use Hadwiger-Nelson: ask if, for her $x$, there is never a plane coloring in $3x$ colors such that there are points unit distance apart with the same color. If she says yes, we know $3x > 7$, so $x = 3$; if she says no, we know $3x < 4$, so $x = 1$; and if she is uncertain, we know $4 < 3x < 7$, so $x=2$. – Chris Oct 04 '13 at 21:07
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    I like this, but what if she has solved the twin primes conjecture, and simply hasn't published yet?! The question does say she "knows EVERYTHING about these numbers" – Joe K Oct 04 '13 at 21:21
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    if she had solved the twin primes conjecture and she did not rush to publish it, deciding instead to ask a riddle, she would be really weird... – mau Oct 05 '13 at 12:07
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Question 1:

I'm thinking of a huge integer number with last digit $7$.

Is this ratio $~~\dfrac{\mbox{my number}}{\mbox{your number}}~~$ integer?

  • $1$ (yes)
  • $2$ (no)
  • $3$ (I don't know)

Question 2:

Consider series $\displaystyle\sum\limits_{n=1}^\infty a_n, \quad (a_n>0, ~~~ n \in \mathbb{N})$, such that there exists limit (see Ratio test) $$ L = \lim_{n\to \infty} \frac{a_{n+1}}{a_n}. $$ If $L+1$ is equal to your number, is this series convergent?

  • $1$ ($L=0$, yes)
  • $2$ ($L=1$, I don't know)
  • $3$ ($L=2$, no)
Oleg567
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The question doesn't state that the BOY is an expert in maths, so he'd probably go for:

Please say "no" if your number is 1, "don't know" if your number is 2, or "yes" if your number is 3.

Grimm The Opiner
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This was suggested by a friend:

If $k$ is your number, does $\mathbb{S}^{3k-2}$ have an exotic smooth structure?

Neal
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    Is there a prize for the answer which is both most esoteric and most poetic to the uninitiated? This must surely be a contender... – dardisco Oct 05 '13 at 03:55
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    I hope the girl really does know everything about maths, otherwise the answer would be "Have you been smoking weed again?" – Alois Mahdal Oct 06 '13 at 16:41
  • Can anyone explain this one? – YoTengoUnLCD Jul 15 '16 at 03:31
  • @YoTengoUnLCD It is known that $\Bbb{S}^1$ has exactly one smooth structure (inherited from $\Bbb{R}$), that $\Bbb{S}^7$ has many exotic smooth structures, and unless something has changed in the last three years, it is not known whether $\Bbb{S}^4$ has any exotic smooth structures. Thus, "no" $\rightarrow k = 1$, "I don't know" $\rightarrow k = 2$, and "yes" $\rightarrow k = 3$. – Neal Jul 15 '16 at 12:37
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Let $n$ be the number you are thinking. And let $x$ and $y$ be positive integers I am thinking. Is there a positive integer solution $z$ for the following equation?

$$ x^n+y^n=z^n $$

  • Yes then $n=1$
  • I don't know then $n=2$
  • No then $n=3$ because of the following proven conjecture

It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvelous proof of this, which your browser is too narrow to contain.

kiss my armpit
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  • FLT is proven and even otherwise n=2 has solutions ( like a conic?) – ARi Oct 04 '13 at 20:22
  • @ARi: What is wrong? – kiss my armpit Oct 05 '13 at 03:29
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    The downvote for this one was unfair! This is a good one. @ARi, $x$ and $y$ are unknown to the girl, so if $n=2$ there may or may not be a positive integer $z$ that satisfies the equation. The poster knew FLT is proven, so $n=3$ gives no solutions. The case $n=1$ will always have solutions. – Bennett Gardiner Oct 05 '13 at 04:12
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If $x$ is your number, is my brother's height in metres more than $10(x-2)+2$?

TonyK
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    it returns "no" "no" "i don't know" ? – Gintas K Oct 03 '13 at 09:30
  • I didn't quite get it right. I think it's correct now. – TonyK Oct 03 '13 at 09:34
  • I have a brother, is his height in centimeters > x^7? could be good one :) – Gintas K Oct 03 '13 at 09:41
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    This answer shows that you do not have to be a professional in math to solve this question (+1). I will ask my wife and kids. – drhab Oct 03 '13 at 09:42
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    Just figured a nice question: There is the sequence of phrases "1.Yes, 2.No, 3.I don't know", which word would your number point to? :DDD – Gintas K Oct 03 '13 at 09:50
  • @GintasK that's tricky and I like it, but in programming-speak, the return value of the question is a bool? type, but this is a string type that must be implicitly cast. – xdumaine Oct 03 '13 at 16:16
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    One possible answer: "Trick question, you don't have a brother." – Joe Z. Oct 04 '13 at 04:58
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‘Is there a perfect number $n$ such that $n+1$ is a multiple of your number?’ If her number is $1$, the answer is obviously yes; if her number is $2$, the answer is I don’t know, since it’s not known whether there are any odd perfect numbers; and if her number is $3$, the answer is no, because every even perfect number greater than $6$ is congruent to $1$ mod $3$.

Brian M. Scott
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    +1. But ... What if the girl's own (as-yet-unpublished) research has decided the Odd Perfect Number issue? – Blue Oct 03 '13 at 09:41
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    @Blue: We know that it hasn’t: it hasn’t been published and independently verified! – Brian M. Scott Oct 03 '13 at 09:43
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    If the girl has decided the OPN issue for herself (especially if she found an example of an odd perfect number), then she won't answer "I don't know" because she thinks she knows. (Of course, it's also possible that the girl had just measured @TonyK's brother's height as part of an anatomical statistics study, so his question isn't necessarily fool-proof, either.) – Blue Oct 03 '13 at 09:53
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    @Blue: Irrelevant: we’re clearly intended to assume that her knowledge represents the state of the art. You’re basically just denying the working assumptions underlying the question. – Brian M. Scott Oct 03 '13 at 09:55
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    @BrianM.Scott - What if there's an odd perfect number? Need that be congruent to $1$ mod $3$? – John Gowers Oct 03 '13 at 12:52
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    That's very like the first solution I thought I'd found: "Does there exist a perfect number that is indivisible by your number?" But then I read the puzzle again - the girl knows EVERYTHING about these numbers, and as such, knows whether any odd perfect number exists even if nobody on our side of the fourth wall does. – Stewart Oct 04 '13 at 23:39
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    @Donkey_2009 This justification ought to be included in the answer. If $n$ is $2 \pmod 3$ then some prime $p \equiv 2 \pmod 3$ must divide $n$ with odd exponent. In this case it is easy to verify that $3 \mid \sigma(n)$, so that we can't have $\sigma(n) = 2n$. – Erick Wong Mar 02 '16 at 00:58
  • @ErickWong Very nice! – John Gowers Mar 02 '16 at 20:11
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The girl take a number in {1, 2, 3}. I say to her: "Ok, now, imagine I know your number and pick one of the other, is your number greater than mine?"

  • If she has picked 1, she will answer "no" because 1 < 2 and 1 < 3.
  • If she has picked 2, she will answer "I don't know" because 2 > 1 but 2 < 3.
  • If she has picked 3, she will answer "yes" because 3 > 1 and 3 > 2.
Maxime
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Is it correct that your number equals 1 or it equals 2 and {any question she don't know answer for}?

Yep, it's a boring answer, but it always works for such kind of problems.

alex
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    Some what ambiguous in written language. Of course you must mean (n == 1) or (n==2 and x) not (n == 1 or n==2) and x – Martin Smith Oct 03 '13 at 12:41
  • @MartinSmith, yep, I don't really know how to specify parentheses in natural language. -.- – alex Oct 04 '13 at 09:00
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Here's my wife's solution:

The boy asks: "I have an equation of the form $ax^2+bx+c=0$ in my mind, in which $b^2-4ac \geq 0$. Is the number of its real roots less than your number?"

  • If the girl's number is 3, then her answer is "yes".
  • If the girl's number is 2, then her answer is "I don't know".
  • If the girls' number is 1, then her answer is "no".
Behzad
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  • 2<2 is a false statement so if her number is 2 she would say 'no', – Meadara Jul 27 '16 at 13:02
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    @Meadara Suppose that her number is 2. She can not know whether $b^2-4ac$ is $0$ or positive. In the former case, the equation has $1$ real roots, and in the latter one it has $2$ reel roots. So the number of real roots can be either equal or less than her number. Hence she just can answer "I don't know". – Behzad Jul 28 '16 at 16:16
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"If two less than your number is the second derivative of a function at a turning point, is that point a local minimum?"

Glen O
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    "no", the second derivative is zero does not imply the point is a local minimum.

    For this to work you have to say: "I'm thinking of a function with a point (x,y) where the derivative is two less than your number. Is that point a local minimum?" Of course this may be problematic since you don't know her number and she knows you don't know her number so she should know you can't really be thinking of such a function. In that case the correct answer is "you're lying"

     – dspyz Oct 03 '13 at 19:08
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    The second derivative being zero doesn't imply that it's a local minimum, but it can be a local minimum with the second derivative being zero (see $x^4$ at $x=0$). Hence her answer would be "I don't know". – Glen O Oct 04 '13 at 03:41
  • The answer is "no". "No, not 'If two less than your number is the second derivative of a function at a turning point, that point is a local minimum'". If you don't say otherwise, it's assumed that free variables in an expression (in this case, the variable being "a function") are universally quantified, not assigned to unknown values. That's why the "I'm thinking of" part is so important. – dspyz Oct 07 '13 at 22:17
  • @dspyz: I'm sorry, but you're wrong. If the second derivative of a function at a turning point (or, more generally, a stationary point) is zero, it CAN be a local minimum, or it can be a local maximum, or an inflection point. Without a defined function, there is insufficient information to determine the nature of that point with regards to maximum/minimum. Let me put it this way: the exact response, if it weren't restricted, would be "It might be, but I'd need to see the function, first" - in other words, she doesn't know the specifics necessary to answer the question. – Glen O Oct 08 '13 at 01:17
  • You didn't give a function. If you said "I'm thinking of a function for which two less than your number is the second derivative at a turning point...". In that case the answer would be what you say. You said "if two less than your number is the second derivative of a [meaning any] function at a turning point, [then] is that point a local minimum?", to which the answer is "no". "I don't know" is incorrect because there's nothing here she doesn't know. Perhaps if you said "Given an unknown function..." she'd say "I wouldn't know", but that's not the same thing. – dspyz Oct 08 '13 at 16:46
  • As I didn't specify the specific function, the question is clearly referring to an arbitrary function. The information provided in the question is insufficient to determine the answer, and thus the response would be "I don't know". You seem to think that specification of the function is inherent in saying "a function", and this confuses me. You're saying her answer would be "no"... but if I then said "the function is $x^4$, you're a liar", she's have to admit to having lied. You are WRONG. Just admit it. – Glen O Oct 09 '13 at 08:22
  • This argument has gone on far too long and is probably just a semantic issue. Please rephrase your query about the second derivative in predicate logic. If the answer for 2 is still "I don't know" (and it appears to be the same query), then I'll believe you. – dspyz Oct 10 '13 at 15:28
  • To my knowledge, predicate logic doesn't have an "I don't know" response. But let's try something a little more formal, by establishing "I don't know" to be equivalent to "I do not currently have enough information", as this is something we can treat a little more formally. Now, the formal way to ask the question, using a bit of extra specification for clarity, is "There is an even-order polynomial with one real turning point*. You are provided the information that the second derivative is X at this point. Is this function bounded below?" – Glen O Oct 11 '13 at 04:47
  • Yes, but where does this polynomial come from? If the boy knows what the polynomial is, then he already knows the girl's number by virtue of the derivative being X at that point. If the girl knows what it is, then the "I don't know" response is inappropriate. If neither of them know what it is, then the polynomial is hypothetical and the correct response is "I would not have enough information" (eg. "I wouldn't know"), rather than "I don't currently have enough information". – dspyz Oct 13 '13 at 21:39
  • The information defining the requisite feature of the polynomial is held by the girl, but she does not know the whole polynomial. The definition itself is held by the boy. It is not some hypothetical polynomial, as the necessary trait restricts the features of the polynomial. It's like saying "the sum of two odd squares will always take the form $2n$ for some odd integer $n$" - this is true irrespective of the specific squares. The squares might be hypothetical, but the trait being examined is not. – Glen O Oct 17 '13 at 04:25
9

I get the sense that there's a classic comedy routine in here somewhere.

"You know, they give baseball players these days very peculiar names. [...] Well, now, let's see ... On our team we have: NO's on first, YES's on second, and I-DON'T-KNOW's on third ..."

Blue
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9

Does the girl know computing addition to math? Probably.

"Here is an array of three character strings, indexed from [1]:

s[] = { "Yes", "No", "I don't know" }

what is the value of s[x] where x is the number you are thinking of?"

Basically, the space of three possible answers can be used as symbols to encode the information directly.

Justification, in the light of comments:

The other answers differ in that they employ an arithmetic and logical coding trick: arithmetic is applied and then logic to produce an answer, whose truth value or in determinacy is then rendered to English "Yes", "No" or "I don't know".

It is just as valid and "mathematical" to simply obtain these symbols directly without using arithmetic coding.

Furthermore, it can still be regarded as arithmetic coding, because the answer strings are made of bits, and can therefore be coded as numbers: for instance, the bit patterns of the ASCII characters can be catenated together and treated as large integers. s is then effectively just a numeric table lookup which maps the indices 1 through 3 to integer symbols which denote text when broken into 8 bit chunks and mapped to ASCII characters.

A lookup table, though arbitrarily chosen, is a mathematical object: a function.

Furthermore, the displacement calculation to do the array indexing is arithmetic; we are exploiting the fact that the information we are retrieving is numeric and can be used to index into a table. Otherwise we would have to specify an associative set relation instead of a function from the integer domain. ("Here is a mapping of your possible state values to the symbols I'd like you to use to send me the value.")

This answer reveals that the question is basically uninteresting. An entity holds some information that can be in one of three states, and there is to be a three-symbol protocol for querying that information. It boils down to, give me the symbol which corresponds to your state, according to this state->symbol mapping function. I would therefore argue that the convoluted arithmetic coding is the hack answer not this straightforward coding method. In computing, we sometimes resort to arithmetic encoding hacks when we have to use a language that isn't powerful enough to do some task directly, or simply when the resources (time, space) aren't there for the cleaner solution.

Kaz
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  • @Kaz this is kinda hacking the answer! I don't think it is wrong, i just believe that a mathematical second option would complete your answer :) – chubakueno Oct 04 '13 at 16:26
  • @kaz There is a level of in correctness as you are to ask a decidability problem which may be referencing the system state ( which is the girl's number); but here you ask a functional problem . Notice the difference. – ARi Oct 04 '13 at 20:27
  • @kar But when I think about it ever functional problem can be broken down into a series of Decision problems – ARi Oct 06 '13 at 16:07
  • Only thing that is left is to refactor the whole riddle to use "one", "two" and "three" instead of "Yes", "No" and "I don't know". Then the quest is complete: the question is now absolutely uninteresting. Math, meet CS; CS, meet Math. – Alois Mahdal Oct 06 '13 at 16:53
  • @AloisMahdal Exactly. So the question needs constraints to avoid this. Like: you cannot "program" that person to just map the numbers to answers with custom functions. The answering person only gives tri-state boolean answers based on evaluating purely arithmetic questions in which the secret number can appear as an operand, and knows standard operations like exponentiation, logarithm, sine, cosine, ... – Kaz Oct 06 '13 at 17:16
  • You really cannot program the person? First, you have been told that the person has installed "maths" library, and assuming that also "language" library and also some kind of programming language interpreter. Then you send your code snippet in and wait for reply. Aren't you effectively programming the person? – Alois Mahdal Oct 06 '13 at 17:33
  • @AloisMahdal NO, you are just using the person's standard library (ISO defined and everything). :-) – Kaz Oct 06 '13 at 17:39
  • Anyone can code the sledge hammer three options question. The answers that use mathematical tricks require mathematical understanding from the questioner. Why is that not a useful problem to challenge a student's understanding of certain mathematical concepts. So, I disagree, it's not an uninteresting problem. There is still some mathematical art going into many of the solutions being presented here. And there is at least one which is quite elegant. – Geoff Pointer Oct 11 '13 at 00:05
8

The boy gives the three numbers different names: "Yes", "No" and "I don't know". So the question is, "what is the name of your number?"

Xiaochuan
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8

I am thinking of a positive integer. Is your number, raised to my number and then increased in $1$, a prime number?

$$1^n+1=2\rightarrow \text{Yes}$$ $$2^n+1=\text{possible fermat number}\rightarrow \text{I don't know}$$ $$3^n+1=2k\rightarrow \text{No}$$

chubakueno
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6

$\frac12-it$ is a zero of the zeta function. Is $\frac{N-1}{2}\frac12+it$ also a zero of the zeta function?

$N = 1 \implies$ no, since $it$ is not on the critical strip.

$N = 2 \implies$ I don't know, since the Riemann hypothesis has not been proved.

$N = 3 \implies$ yes, since the conjugate of a zero is another zero.

Daniel R
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  • Does the negation of the Riemann hypothesis really mean that $\frac14+it$ can be a zero of the zeta function whenever $\frac12-it$ is a zero of the zeta function? – John Gowers Oct 03 '13 at 12:30
  • @Donkey_2009 Good point, I don't know. Has it been proved that $\frac14 + it$ can/cannot be a zero if $\frac12+it$ is? – Daniel R Oct 03 '13 at 12:58
  • Well, it certainly hasn't been proved that it can, since that would negate the RH. My guess is that it's probably known that that sort of thing isn't possible (quite a lot is known about the zeta function), but I couldn't say for sure. – John Gowers Oct 03 '13 at 13:11
  • @Donkey_2009 Yeah, I edited to add the "can" part. Let's say it was a temporary (?) brain lapse. Regarding "cannot", you are probably right it may be known. If so, my answer bogus. – Daniel R Oct 03 '13 at 13:23
  • I think maybe your edit didn't go through or something, because I can't see it ;-) – John Gowers Oct 03 '13 at 13:58
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    @Donkey_2009 The edit was in my comment. :) – Daniel R Oct 03 '13 at 14:00
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    This fails because there's nothing that prevents a hypothetical question answerer from having a proof or disproof of it. In fact, the problem states that the girl "knows EVERYTHING about these three numbers". – Olathe Oct 04 '13 at 20:05
6

Two silly, brute-force, examples, which (I hope) give two fairly extensible ways of constructing an answer to this problem. $n$ refers throughout to the girl's number.

The unsolved problem in mathematics approach: (E.g., @alex's answer)

Is there a Moore graph of girth $5$ and degree $f(n)$? Here,

$$ f(n)=\begin{cases}2&n=1\\4&n=2\\57&n=3\end{cases} $$

$2\mapsto$Yes (the Petersen graph)

$3\mapsto$No (A Moore graph of girth $5$ may only have degree $2,3,7$ or $57$)

$57\mapsto$I don't know (It is unknown whether a Moore graph of girth $5$ and degree $57$ exists).

The unconditional (on any unsolved result being unsolved!) probabilistic approach: (E.g., @Ben Millwood's answer)

Maybe the girl has constructed a Moore graph of girth $5$ and degree $57$ in her head, or read a recent paper exhibiting such a graph. In that case, the following approach works.

Let $G(n)$ be a random variable taking values in the set $\{1,2\}$. The probability distribution is defined as follows: $G(1)$ is always $1$, $G(2)$ is always $2$ and $G(3)$ is $1$ with probability $0.5$ and $2$ with probability $0.5$.

[Fix a point in the sample space.] Is it the case that $G(n)=1$?

John Gowers
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6

Most of the answers seem to rely on relatively complex mathematics, that I couldn't easily do in my head. I'm hoping this is the simpliest answer out there, or at least, one of the simplest.

If I take your number, and subtract 2 from it, then take the reciprocal, is it positive? In other words: $\frac{1}{x-2}$

  • Yes- Number must have been 3
  • No- Number must have been 1.
  • I don't know- Is 1/0 positive or negative? It could be either, and thus I Don't Know is the appropriate answer.
PearsonArtPhoto
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  • @Kevin: There's a reason that I participate here and not on English. Spelling... Sigh. – PearsonArtPhoto Oct 03 '13 at 15:23
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    The answer for 1/0 isn't an "I don't know", it's "no". It's not positive (it's also not negative) – dspyz Oct 03 '13 at 18:56
  • @dspyz: That is one specific definition of infinity. You can also define it using a one-sided limit. – Ryan Oct 04 '13 at 05:42
  • @Ryan, division is already defined algebraically, and dividing by zero is not infinite. It has no result, which means the result isn't positive. The hypothetical question answerer making up a completely new definition is not really valid. – Olathe Oct 04 '13 at 20:03
  • @Ryan Infinity can be a one-sided limit, but in that case 1/0 is not infinity because it's not defined. In general, when 1/0 is defined (such as on the Reimann Sphere), it's not negative or positive. – dspyz Oct 07 '13 at 22:25
6

How about this one:

Let's say your number is $n$.
For every even number $x$ ($x>2$) is that true that $x+n$ is representable as a sum of $n$ primes?

Kostya
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5

The mathematician's answer:

I am thinking of a function $f:\{1,2,3\}\to\{0,1\}$. $f$ takes $1$ to $0$, $2$ to $1$ and $3$ either to $0$ or $1$, but I'm not telling you which.

Let $n$ be your number. Is $f(n)$ equal to $1$?

John Gowers
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5

If k is your number, does each continuous mass distribution µ in $\mathbb{R}^{3k-2}$ admit an equipartition by hyperplanes ?

Minuteman
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4

Here's one involving probabilistic trials.

If I were to choose a random variable $x$ distributed at $N(2, 0.01)$ but cut off at $[1.5, 2.5]$, would the number you're thinking of be higher than $x$?

If you're thinking of $1$, your answer is "no".

If you're thinking of $2$, your answer is "I don't know", since there's a 50-50 chance either way.

If you're thinking of $3$, your answer is "yes".

Joe Z.
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4

Assume the girl's number is X.

The boy asks:

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

Terry Li
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4

"Using the number of characters in the first word of you possible answers (Yes, No, I don't know), which options length corresponds to your selected number?"

eskimo
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    I would rephrase as "If the number of characters in the first word of your answer must be equal to your selected number, what would you answer?", given the girl is only allowed to say the three answers. Good solution +1 – peterwhy Oct 04 '13 at 14:17
3

For an unknown $x$, is your number 1 or (2 and $x$)?

YES means that the number is 1
NO means that the number is not 1 and not 2 -- so it's 3
I DON'T KNOW means that the number is 2 (since the (2 and $x$) clause will be true if $x=2$ and false otherwise)

SamYonnou
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3

Without any math involved:

Boy: I'll map your numbers to your answers. 1 means "NO", 2 means "YES", 3 means "I don't know", the first word that comes out of your mouth, will you please say the phrase that is mapped to the number you are thinking of?

Girl: ahhhhhhhhhh
Boy: No, you have to say it

First dimension: 

Girl: NO
BOY: Then it's 1

Alternate dimension:

Girl: YES
BOY: Then it's 2

Altnerate dimension after that:

GIRL: I DON'T KNOW
BOY: Then it's 4, I mean 3
sksallaj
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    She can answer "no" as in "no, I will not answer the phrase that is mapped to the number I am thinking of", and be answering the question correctly and honestly. – Beska Oct 03 '13 at 20:41
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    That's an existential argument, we're not talking about liars here or someone who thinks that is an unfair approach.. we're talking about how to deal with the asking a question once, and getting the answer from it. She could very well lie if she wanted to, or she doesn't have to answer the question at all, whether or not the boy is asking "fair" questions.. – sksallaj Oct 04 '13 at 01:07
  • I disagree. She is asked to "answer a question". Presumably, correctly and honestly. I don't see anything in the original question that states she has to be willing to ignore the rules of normal conversation to relay the answer in code. – Beska Oct 04 '13 at 12:08
  • The point of the question is to discover how to discover the answer without this kind of work. Otherwise I could solve the riddle with only one possible word response, and an infinite number of possible answers, and do so trivially, by saying "answer 1 corresponds to a 1 second delay before your response, answer 2 to a 2 second delay before your response, etc." It's bending the original intent of the question. – Beska Oct 04 '13 at 12:11
  • But that's from your perspective of the riddle, not anyone else's, that doesn't have to be part of what the riddle actually is asking. The riddle is really generalized, because the boy asking the question really doesn't know what's going on in the girl's head. The only way this riddle works is if the girl answering is honest and open and willing to compromise.

    How do you know what the point of the question is? It doesn't mention it anywhere in the riddle that you have to use math or to go about it in a certain way. It said, "One question" and the boy knew the answer afterwards.

     – sksallaj Oct 04 '13 at 18:55
  • The mapping phrase the boy mentioned isn't a question. The goal of the riddle.. ask one question, if the girl says, "yes", "no", or "i don't know", the boy will know the answer afterwards. That's it. No other interpretation, no limit or bounds to go about it. You are given one question about the number in the head of the girl and my answer does exactly that.

    Just because you think it's too simple or because it's too much work doesn't take away from the fact that it does exactly what the riddle asks us to do.

     – sksallaj Oct 04 '13 at 18:58
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    Your one question is whether she'll do something. Answering whether she'll do something or not is not the same thing as using your code. – Olathe Oct 04 '13 at 20:12
3

"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 $n$?"

peterwhy
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1

"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 $n = \mathrm{TREE}(3)$.

r.e.s.
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  • 1
    While its precise value may be unprovable, you may have to work harder to demonstrate that its parity is not provable. – Ben Millwood Oct 04 '13 at 11:30
1

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".

Bennett Gardiner
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  • 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...) – Asaf Karagila Oct 08 '13 at 07:45
  • Sure, but I think the overwhelming consensus in this thread is that she has all current day knowledge but is not some omniprescient being. – Bennett Gardiner Oct 08 '13 at 12:31
  • 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? :-) – Asaf Karagila Oct 08 '13 at 12:38
  • Nope! Read them again, the girl needn't know what number you are thinking of. – Bennett Gardiner Oct 09 '13 at 05:37
0

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

Rigel
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0

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.

Jack D'Aurizio
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0

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...)

DanielV
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0

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.

shrinath
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  • 2
  • 4
0

If there is no generally agreed "$>$" operator for complex number field: "Is $e^{\left(i\frac{n-1}{2}\pi\right)}>0$?"

peterwhy
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