# You have one question to tell whether the number I'm thinking of is 1, 2, or 3

This is an interesting puzzle which was passed to me by a friend some time ago. I do know the answer, but will refrain from self-answering on this to see where it goes.

I'm thinking of a number: 1, 2, or 3. You may ask me one question, which I will answer to the best of my ability. I may not, however, tell you my number or any codified version of my number. I can only answer yes, no, maybe, I don't know, etc.

How do you tell the number I'm thinking of?

• This question was posted on Math.SE a while back. – Joe Z. May 21 '14 at 17:56
• Everytking boils down to asking a question, which may give three different answers - yes, no or maybe/don't know. – Spook Jul 17 '14 at 10:12
• "any codified version of my number" sort of includes /any/ answer that can uniquely identify the number. – perfectionist Oct 7 '14 at 13:50
• @perfectionist What I meant by that was asking something like "What is your number plus three?" or "If 1 is A, 2 is an, and 3 is C, what is the matching letter for your number?" - in other words, the "spirit of the problem," in a sense. I agree it isn't the best way to write it, but... I'm not sure how else to. – user20 Oct 7 '14 at 13:59
• I tried this on a friend, and her 'question' for me was "What is the number you're thinking of." She doesn't understand these puzzles. – corsiKa Nov 3 '14 at 18:47

Would it work if I ask:

"I'm thinking of a number: either 0 or 1. Is the sum of our numbers greater than 2?" I could figure out the answer if that's a question I'm allowed to ask.

If your number is 3, then you would say "yes" because no matter what you add, it would be true.

If your number is 2, you would say "I don't know". If my number was 0, then the answer would be no, but if my number is 1 the answer is yes. So you don't know.

If your number is 1, then you would respond "no". No matter what number you add, the number will never be greater than 2.

• scenario 2 could also be answered with "maybe" – DLeh Jun 5 '14 at 15:16
• @DLeh One could answer "maybe" to any question, except that the conditions say "to the best of my ability". – Jim Balter Jul 17 '14 at 9:22
• I have upvoted this answer... but somebody else please upvote. Turn that $99$ into $100$ :D – Mr Pie Sep 22 '18 at 11:50
• Isn't this exactly the same as [this]math.stackexchange.com/a/513280/561119? – PotatoLatte Sep 22 '18 at 14:14
• um how do you do a link again? – PotatoLatte Sep 22 '18 at 14:14

If you flip that many coins, will you get two the same?

Because

As Doorknob put it... "1 => no, because it's only one, 2 => maybe, because it could be one of TH, HT, TT, or HH, 3 => yes, because a coin flip has only two possible outcomes"

• +1 not for originality but because this is the the most elegant solution here. No trickery, easy to understand. – ThePopMachine Jul 11 '14 at 0:38
• Agreed, a very elegant and most efficient solution. – Michael Rize Feb 15 '15 at 11:59
• Probably "at least two the same" just to cut out trickiness on the part of the answerer. – thumbtackthief Apr 13 '15 at 17:40

Here is my answer:

I am thinking of an odd number , does your number divide the number I am thinking of?

Because

1 -> yes
2 -> no
3 -> maybe

• This is the most elegant and understandable answer to me. – Philipp Aug 12 '18 at 10:42

Here's my thought process on this:

• The three answer possibilities must be "yes," "no," and... something else.
• Perhaps there is a way to get an answer of "maybe" or "sometimes?"
• Therefore, this would have to involve some kind of randomness.
• The question should be in the form of "Either X or Y is true. Is your number {some condition}?"
• A simple condition I can think of is "even or odd."
• Is there a way to get a known even and known odd result for two of the numbers, and an unknown one for the other?
• Let's make a little chart:

1 2 3
O E O n^2
E O O floor(n/2)
O O E ceil(n/2)
E E O floor(n/3)
O O O ceil(n/3)

• Hey, look! For floor(n/2) and floor(n/3), our conditions are satisfied!

• So, my final question is:

I will divide your number by either 2 or 3 and round it down. Is the result even or odd?

If the answer is "even," the number was 1. If it's "odd," the number was 3. If it's "I don't know," the number was 2.

• Note that an alternate question could use n^2 and floor(n/3) instead (I will either square your number or divide it by three and round down).
• 0 is not an even or odd number... – awesomepi Mar 18 '15 at 20:31
• @awesomepi 0 is an even number. How could a number be neither even nor odd? – Doorknob Mar 18 '15 at 22:45
• @Doorknob $\frac{1}{2}$ is neither even nor odd. ;) – Ian MacDonald Apr 21 '15 at 13:30
• @IanMacDonald ... okay, fine, all integers must be even or odd. :P – Doorknob Apr 21 '15 at 13:31

Slightly tongue-in-cheek, but a variation on the "maybe" option provided the other player interprets it literally enough.

Keep evaluating powers of your number until you reach one that exceeds 6. Is it an odd power? Answer yes = 2, no = 3, infinitely long pause = 1 …

As is the case with all information puzzles, we have to find away to map each outcome (yes, no, I don't know) to a different number (1, 2, 3). To do this, we need some property that's unknown for exactly one of these three numbers.

There are multiple ways to do this. The most basic (and most reliable) way would be to introduce a number that only you know so I wouldn't know:

I'm thinking of either 1.5 or 2.5. Is your number greater than mine?

1 is smaller than either of these numbers, so the answer will always be no. 3 is greater, so yes. And 2 could be either greater or smaller, so I don't know.

Another way, which is less reliable, is to introduce an unknown is to use some unknown property in mathematics:

Is TREE(your number) divisible by your number?

The Kruskal tree theorem produces a sequence of numbers such that TREE(1) = 1 and TREE(2) = 3, but TREE(3) is so astronomically large that it makes Graham's number look like epsilon in comparison. Moreover, it's not even known how to calculate it, so we cannot even tell whether it's even or odd.

In this case, if I'm thinking of 1, the answer is yes, if I'm thinking of 2, the answer is no, and if I'm thinking of 3, I'm certainly not going to know whether TREE(3) is divisible by 3. The reason this one is less reliable is because if we ever do find a way to calculate TREE(3), the uncertainty will no longer be there.

Yet a third way, the least reliable of all, is to use some form of contingent unknown, like a form of obscure trivia or some event that has a relatively unpredictable or unknown occurrence.

Is the current world population greater than (your number + 5.3) billion?

Will Easter Sunday ten years from now happen less than (your number * 32) days after February 1?

Is the last bit of the SHA-256 hash of "entanglement" strictly smaller than (your number - 1)?

The current world population is currently between 7.2 and 7.3 billion. It reached 6.3 billion ten years ago, and won't reach 8.3 billion for another ten years, but somebody who didn't actually look up the population tables wouldn't know this fact for sure.

Easter Sunday is defined as the Sunday after the first full moon after the vernal equinox, so it can't happen before March 20 in any year (which is the earliest date the vernal equinox can happen), and it can't happen after April 25 (36 days after the vernal equinox) either. So it's not going to happen before March 3 or after May 8 either. But the date in the middle, April 5/6, is right in the middle of possible dates for Easter, so unless you've memorized the Computus table, it's not generally feasible to predict what day Easter is going to fall on.

The SHA-256 hash of "entanglement" is generally impossible to calculate in a few seconds in your head, unless you looked it up ahead of time, but you do know that its last bit (which is either 0 or 1) will always be less than (3 - 1), and never less than (1 - 1).

Although this method isn't completely reliable in that they might know that piece of trivia, the key is to provide two completely ridiculous values on each side and one that can be argued either way in the middle that the average person would answer "I don't know" to.

For the record:

The current world population at the time of posting is about 7.23 billion, which is less than 7.3 billion. This makes the answer "no" when my number is 2.

Easter Sunday will happen on March 31 in 2024, which makes the answer "no".

The SHA-256 hash of "entanglement" is 961B164F23EB33F8FDA12C95E8BD93F6 32A08A8B8A0A18B3DDE1CFE8926875FF, which has a last bit of 1. So the answer is "no".

• How about something like "does your number as a digit 10^100 consecutive times in the base-3 representation of pi before the digit "1" appears 10^100 + 1 consecutive times? Certainly true for 1 and false for three. It's been mathematically proven that every possible sequence of digits in any finite base appears somewhere in pi, so for "2" the answer would be well-defined, but I can't think of any way one could determine the answer with any present or imaginable future technology absent some great leap in mathematics. – supercat Jun 8 '14 at 17:36
• How do you know it's false for 3? – Joe Z. Jun 8 '14 at 21:11
• The base-3 representation of pi consists entirely of zeroes, ones, and twos. – supercat Jun 8 '14 at 21:22
• Oops, forgot you were talking about base 3. Ignore me. – Joe Z. Jun 8 '14 at 21:27
• "it's not even known how to calculate it, so we cannot even tell whether it's even or odd" - a very dubious statement. Let $n$ be the largest Mersenne prime if there is one, $n=1$ otherwise. Then it is not known how to calculate $n$, but it is definitely odd. – David Aug 26 '14 at 12:09

Here's my simple answer:

What word has that many letters? Please answer with yes or no.

Example:

- Yes -> 3
- No -> 2
- Er... I can't. -> 1

• I like it. It's asking for a yes or no answer, but not for the meaning behind them... – Brian Minton Sep 9 '17 at 17:26

The question I would ask is this (where $n$ is the number you are thinking of):

Is $\lim_{x \to n} \frac{1}{x-2} > 0$ (where we are approaching $n$ from either the left or the right)?

The possible answers are:

• If $n = 1$: $\frac{1}{1-2} = -$, so the answer is "No".
• If $n = 2$: $\lim_{x \to 2^+} \frac{1}{x-2} = +\infty$ but $\lim_{x \to 2^-} \frac{1}{x-2}$ = $-\infty$ so the answer is "I don't know".
• If $n = 3$: $\frac{1}{3-2} = 1$, so the answer is "Yes".
• The answer to “is $\lim_{x\to2} 1/(x-2) \gt 0$” is not “I don't know” (assuming the answerer is a truthful mathematician), but “it is undefined”. “No” may be a valid answer too, depending on how the question (which is formulated in English) is interpreted as a mathematical statement (“is the proposition $\lim_{x\to2} 1/(x-2) \gt 0$ true?” → no, but “what is the truth value of the proposition $\lim_{x\to2} 1/(x-2) \gt 0$” → undefined). – Gilles 'SO- stop being evil' May 22 '14 at 20:33
• @Gilles Can you not apply the same reasoning to any of the other answers to this puzzle? That limit in the $n=2$ case is either $+\infty$ or $-\infty$, so the answer to "is it positive" is not known -- you can call it whatever you'd like but that's just a question of semantics. I don't really see what the problem is. – arshajii May 22 '14 at 20:40
• @arshajii No, the limit is not “either $+\infty$ or $-\infty$”. There is no limit. – Gilles 'SO- stop being evil' May 22 '14 at 20:48
• @Gilles Well I suppose I could end the question with "where we are approaching $n$ from either the left or the right" (but I still don't think it's really necessary). Would that suffice? – arshajii May 22 '14 at 20:49

"Are there at least $N + 27$ days in February, where $N$ is your number?"

Because

• if $(N = 1)$ then $(N + 27 = 28)$ --> Yes

• if $(N = 2)$ then $(N + 27 = 29)$ --> Maybe

• if $(N = 3)$ then $(N + 27 = 30)$ --> No

I wanted a solution that avoids uncertainty. So here's a good logical contradiction:

Is exactly one of these statements true?

(a) Your number is not 2.

(b) Your number is 3 and your answer to this question is Yes.

Analysis

If the number is 1, then (a) is true and (b) is false, so the answer is Yes.

If the number is 2, then (a) is false and (b) is false, so the answer is No.

If the number is 3, then (a) is true. Now:

If the answer is Yes, then (b) is true and therefore the answer must be No.

If the answer is No then (b) is false and therefore the answer must be Yes.

By contradiction, neither Yes nor No can be answered, and some other response must be used (e.g. Unanswerable).

Summary

If Yes, then the number is 1.

If No, then the number is 2.

If Unanswerable, then the number is 3.

As you can see in this topic, there are a lot of possible solutions of course.
But there is general solution, which includes all of them. It is:

Are you thinking of 1 OR you are thinking of 2 AND [statement with unknown result]?

The interpretation of the answer is clear:

1 -> yes
2 -> I do not know
3 -> no


My most preferable solution is

Are you thinking of 1 OR you are thinking of 2 AND tomorrow will be rain?

Just because it is simple and it is clear how it works.

But any other solution will be equivalent to given general solution (up to a interchange of numbers).

For example accepted answer of Ice-9:

"I'm thinking of a number: either 0 or 1. Is the sum of our numbers greater than 2?"

is equivalent to:

Are you thinking of 3 OR you are thinking of 2 AND I thinking of 1?

Or another solution of Doorknob:

"I will divide your number by either 2 or 3 and round it down. Is the result even or odd?"

is equivalent to:

Are you thinking of 1 OR you are thinking of 2 AND I thinking of 3?

Even such sophisticated solution, which arshajii gave:

"Is $\lim_{x \to n} \frac{1}{x-2} > 0$ (where we are approaching $n$ from either the left or the right)?"

is equivalent to:

Are you thinking of 3 OR you are thinking of 2 AND $\lim_{x \to 0} \frac{1}{x} > 0$?

"Is there infinitely many n such as n and n + 'your number' -1 are primes?"

Because

1. Yes (Euclid)

1. No, obviously

1. I don't know, still working on twin prime conjecture

• ...until somebody solves that damn conjecture. $(+1)$ :D – Mr Pie Sep 8 '18 at 8:45

I would say:

Considering that I am also thinking of a number between 1 and 2, is you number higher than mine?

1-> No
2-> Maybe
3-> Yes

• I guess you might have to say that $\ast$ (ROT13) lbh ner guvaxvat bs n ahzore orgjrra 1 naq 2 vapyhfvir, sbe gur "znlor" cneg $\ast$ but that is just being technical. $(+1)$ :D – Mr Pie Sep 8 '18 at 8:49
• how did I not think of that – user3453281 Sep 9 '18 at 18:50

Divide 2 by one less than the number you were thinking of. Is the result an even number?

• Yes: You were thinking of 2 (because 2/(2-1) = 2)

• No: You were thinking of 3 (because 2/(3-1) = 1)

• Unanswerable: You were thinking of 1 (because 2/(1-1) = ∞)

I would

write 2.XX on a piece of paper, cover its fractional part with my hand

"Is your number greater than mine?"

Because if

1. Definitely no

2. Maybe (my number would have been $2.00$, so giving no for an answer would not be correct)

3. Definitely yes

We have 1 ball on the table and 0 or 1 balls under a concealed cup. Now I ask you is the amount of balls on the table is larger than the number on your mind?

1. No, because 1 is not larger than the 1 or 2 balls on the table

2. Maybe, because maybe it's larger than 1, but it's not larger than 2

3. Yes, because 3 is larger than the 1 or 2 balls on the table.

• I think this may be confusion over what I mean by "codified" - I'm going to clarify. – user20 May 21 '14 at 14:39
• I hope this answer is more like what you expect, but I still think this is some encoding too :) – martijnn2008 May 21 '14 at 15:01

Soccer World Cup is in progress, so that warrants a themed question (though just a bit late, as group stage is over now - oh well):

Is your number both valid and not less than a number of points that a team got after finishing a group stage match in which they scored at least the same number of goals as their opponent?

A soccer team receives 1 point for a draw and 3 points for a win, 2 is not a valid number of points, so:

1 => yes
2 => no
3 => maybe


I feel this doesn't bode well with the spirit of the posted question as some of the other answers do, though. ;)

• It's valid enough, just not very elegant. – Joe Z. Jun 29 '14 at 3:28

Does the number of letters in your answer match the number you are thinking of?

So if thinking of 1, answer is no.
If thinking of 3, answer is yes.
If thinking of 2, can't truthfully answer yes or no

• Doesn't work, could answer yes or no with 3, both would be true. Which makes that the seed for a new puzzle! – YvesgereY Jul 28 '17 at 9:04

If I pull your number of socks from a drawer with two different pairs of socks in it, will I have a matching pair? 1 - No, 2 - Maybe, 3 - Yes.

Wimbledon is in progress, so let's put that to good use:

Is your number less than a number of sets played in a straight-set win in a Wimbledon tennis match?

Women win in 2 sets, men in 3 sets.

1 -> yes
2 -> maybe
3 -> no


I would

consider a prime number $p\neq 2$

"Does your number divide $p$?"

Because

• If he answers Yes, then the number is $1$.

• If he answers NO, then the number is $2$.

• If he answers MAYBE, then the number is $3$.

• It won't be a "maybe" for the fifth spoiler tag/quote because three divides itself. I have also proposed an edit so as to not spoil a thought/idea for other users attempting to solve (or discover alternative solutions) for the puzzle... but most importantly, welcome to the Puzzling Stack Exchange (Puzzling.SE)! Head here! :D – Mr Pie Sep 8 '18 at 8:41

If you are driving and take your number's worth of turns, (right or left, no off-ramps or U-turns), will you get multiple turns to the same direction? This will work because:

If their number is 1

They will say "no", for obvious reasons.

If their number is 2

They will say "maybe" or "I don't know", because with two turns, you could (where R = a right turn and L = a left turn) do: RL, LR, RR, or LL.

If their number is 3

They will say "yes." Possible turns are: RLL, RLR, RRR, >!RRL, LRR, LRL, LLR, LLL. All of these include multiples of the same direction.

I am thinking of a multiple of 5, but not of 15 - does your number divide mine?

If the number the person is thinking of is 3, it cannot divide my number since I blocked factors of 3, so we get a "no".

If 1, it will always divide my number, since 1 is a factor of every number, so we get a "yes".

If 2, we get a "maybe", because I could just as easily be thinking of 20 as 25, so there is not enough information.

• – xhienne Mar 25 at 17:03

Does any user of puzzling.stackexchange have that many arms?

• If your number is 3, the answer will be "no" (unless we have any aliens on the site)
• If your number is 2, the answer will be "yes"
• If your number is 1, the answer will most probably be "maybe".
• 3 is similarly "maybe" as 1:) – klm123 Jun 8 '14 at 20:37
• @klm123 The answer for 3 is "almost certainly not" whereas the answer for 1 is "probably so". – Jim Balter Jul 17 '14 at 9:33

"Supposing that the number you are thinking of is $n$, if I was to get $n$ parallel one-sided mirrors and face the glass to each other, would I get infinite reflections on each mirror, theoretically?"

Because

1. Nope. There is only one mirror.

2. Yes. Two mirrors can create an infinity mirror if parallel and facing each other.

3. Maybe. It depends on the size. Imagine these lines being one-sided mirrors: $\large\color{blue}{\mid}\:\color{red}{\mid}\:\color{green}{\mid}$.
If the glass side of each blue and red were facing each other, then since they are one sided, there would not be infinite reflections on the glass of green. The same would apply for blue if I turned around red to face green instead, which is facing in the direction of blue and red.
We would need red to be shorter with blue and green facing each other; i.e., ${\large\color{blue}{\mid}}\color{red}{\mid}{\large\color{green}{\mid}}$.
Red can face whichever other mirror (blue or green)... but the question never mentioned the size of the mirrors, so it all depends on that, particularly the size of the red mirror.

would this work?

she is thinking of a number; 1, 2, or 3, and only able to answer yes no or I don't know. I'm given only one question to know her number. so...
"using only 1,2, and 3, if you subtract the number preceding the number you are thinking, and add the number following the one you are thinking, then divide that answer by your number, is the resulting number even?"
the following would happen:
if 1: it would be (1+2)/1 becoming 3/1 simplified into 3 odd: NO

if 2: it would be (2-1+3)/2 becoming 4/2 simplified into 2 even: YES

if 3: it would be (3-2)/3 becoming 1/3 which is a fraction being neither odd nor even so: I DON'T KNOW

this has been in my head for years since my friend told me this and this is the answer I came up with that makes the most sense to me so far.

it's the "I don't Know" part that has me flustered, because that can always be swept under as "NO."

I would ask: If $$N$$ is your number, is $$N-1$$ a natural number which is also an identity element in $$\mathbb{Q}$$ ($$\mathbb{R}$$ or $$\mathbb{C}$$ would also work)?

3: No (2 is neither a multiplicative identity nor an additive one).
2: Yes (1 is multiplicative identity).
1: Maybe (also "I don't know", "It depends" etc.) (0 is additive identity, but some authors don't consider it a natural number).

Does your answer have the same number of letters as is your number numeric value?

A little tricky, question seems not to ask for enough information, however:

answer: NO(2) >> 1 ; answer: YES(3) >> 3 ; answer MAYBE(5)/DON'T KNOW(8) >> 2 (two)

If the number is 2, answers YES (3 letters) and NO (2 letters) are both wrong!

• Oh, in fact almost the same idea as @ Florian F solution, based on contradiction. – z100 Sep 21 '18 at 16:31