# One question to know if the number is 1, 2 or 3

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?"

• '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 3, 2013 at 13:04
• I am thinking of a number greater than 1 and less than 3. Is my number greater than yours? -- That would work. Oct 3, 2013 at 13:50
• @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". Oct 3, 2013 at 16:19
• Fix: I am thinking of a number that is either 1 or 2. Is my number greater than or equal to yours? Oct 3, 2013 at 19:08
• Why not simply I'm also thinking of a number. It's either 1.5 or 2.5. is my number bigger than yours? :) Oct 4, 2013 at 5:36

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

• +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 3, 2013 at 10:45
• 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. Oct 3, 2013 at 15:44
• @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. Oct 3, 2013 at 16:13
• @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. Oct 3, 2013 at 16:40
• 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. Oct 3, 2013 at 17:09

"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 $m$.

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.

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.

• 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. Oct 3, 2013 at 18:45
• @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. Oct 3, 2013 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. Oct 3, 2013 at 19:03
• 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.
– 6005
Nov 10, 2016 at 23:42
• Ultimately the riddle is asking you to SELECT such a bijection $f$ that induces a well-formed "yes/no" question. Oct 20, 2021 at 17:32

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.
• 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. Oct 3, 2013 at 14:12
• I disagree. The correct answer in the 1 case is "I can't do that." Oct 3, 2013 at 17:59
• 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" Oct 3, 2013 at 18:53
• "Is 1/0 a fraction?" The answer is certainly "no", not "I don't know". Oct 4, 2013 at 6:36
• 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. Oct 10, 2013 at 23:40

"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 $\{2, 5\}$.

• 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 $x=2$. Oct 4, 2013 at 21:07
• 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" Oct 4, 2013 at 21:21
• 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 5, 2013 at 12:07

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)

The question doesn't state that the BOY is an expert in maths, so he'd probably go for:

• It is not a question. (just request). Oct 3, 2013 at 11:33
• @Oleg567 what would you reply be if I gave you the command to say "no" if your number is 1, "don't know" if your number is 2, or "yes" if your number is 3? Oct 3, 2013 at 12:06
• Hmm, then complete your sentence with phrase "OK?" Then it will look more as a question :) Oct 3, 2013 at 13:32
• @Oleg567: Then her answer would be "ok" :) Oct 3, 2013 at 14:15
• Or, similarly: If I like the number 1, and dislike the number 2, do I like the number you chose?
– gdw2
Oct 3, 2013 at 16:55

This was suggested by a friend:

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

• Is there a prize for the answer which is both most esoteric and most poetic to the uninitiated? This must surely be a contender... Oct 5, 2013 at 3:55
• I hope the girl really does know everything about maths, otherwise the answer would be "Have you been smoking weed again?" Oct 6, 2013 at 16:41
• Can anyone explain this one? Jul 15, 2016 at 3:31
• @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, 2016 at 12:37

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.

• FLT is proven and even otherwise n=2 has solutions ( like a conic?)
– ARi
Oct 4, 2013 at 20:22
• @ARi: What is wrong? Oct 5, 2013 at 3:29
• 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. Oct 5, 2013 at 4:12

If $x$ is your number, is my brother's height in metres more than $10(x-2)+2$?

• it returns "no" "no" "i don't know" ? Oct 3, 2013 at 9:30
• I didn't quite get it right. I think it's correct now. Oct 3, 2013 at 9:34
• 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. Oct 3, 2013 at 9:42
• 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 Oct 3, 2013 at 9:50
• One possible answer: "Trick question, you don't have a brother." Oct 4, 2013 at 4:58

‘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 $3$.

• +1. But ... What if the girl's own (as-yet-unpublished) research has decided the Odd Perfect Number issue?
– Blue
Oct 3, 2013 at 9:41
• @Blue: We know that it hasn’t: it hasn’t been published and independently verified! Oct 3, 2013 at 9:43
• 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 3, 2013 at 9:53
• @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. Oct 3, 2013 at 9:55
• @BrianM.Scott - What if there's an odd perfect number? Need that be congruent to $1$ mod $3$? Oct 3, 2013 at 12:52

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.

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.

• 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 Oct 3, 2013 at 12:41
• @MartinSmith, yep, I don't really know how to specify parentheses in natural language. -.-
– alex
Oct 4, 2013 at 9:00

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".
• 2<2 is a false statement so if her number is 2 she would say 'no', Jul 27, 2016 at 13:02
• @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". Jul 28, 2016 at 16:16

"If two less than your number is the second derivative of a function at a turning point, is that point a local minimum?"

• "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" Oct 3, 2013 at 19:08
• 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". Oct 4, 2013 at 3:41
• 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. Oct 7, 2013 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. Oct 8, 2013 at 1:17
• 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. Oct 8, 2013 at 16:46

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 ..."

Does the girl know computing addition to math? Probably.

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

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 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 :) Oct 4, 2013 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 4, 2013 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 6, 2013 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. Oct 6, 2013 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 6, 2013 at 17:16

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?"

• An interesting derivative of Peter Olsen's answer. Oct 3, 2013 at 17:23
• ha, I didn't see that one. Oct 3, 2013 at 22:03

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

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

$\frac12-it$ 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.

• 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? Oct 3, 2013 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? Oct 3, 2013 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. Oct 3, 2013 at 13:11
• @Donkey_2009 The edit was in my comment. :) Oct 3, 2013 at 14:00
• 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". Oct 4, 2013 at 20:05

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?

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 $G(n)=1$?

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.
• @Kevin: There's a reason that I participate here and not on English. Spelling... Sigh. Oct 3, 2013 at 15:23
• The answer for 1/0 isn't an "I don't know", it's "no". It's not positive (it's also not negative) Oct 3, 2013 at 18:56
• @dspyz: That is one specific definition of infinity. You can also define it using a one-sided limit.
– Ryan
Oct 4, 2013 at 5:42
• @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. Oct 4, 2013 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. Oct 7, 2013 at 22:25

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 $1$?

• You would have to change it to $f: \{1, 2, 3\} \rightarrow \{$Yes, No$\}$ for it to conform. She can't say 1 or 0, only yes, no, or I don't know. Oct 3, 2013 at 16:12
• @scott_fakename - Good spot! I'll edit it now. Oct 3, 2013 at 16:12

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

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".

Assume the girl's number is X.

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