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Three perfect mathematicians with extremely strong memories are taking an exam. The examiner tells each of them a certain piece of information about $x$ and $y$, which two positive integers between and including 2 and 100 one is divisible by the other:

  • The first one is told the difference $d=x-y$
  • The second one is told the sum $s=x+y$
  • The third one the ratio $r=x/y$.

Then the following conversation takes place:

  • First mathematician: I cannot deduce $x$ and $y$ from their difference $d$.

  • Second mathematician: I knew that you cannot.

  • First mathematician: Don't bother to say that, I already knew that you knew that I cannot. But I'm not sure if you don't know that I know that you don't !!!!

  • Third mathematician: What about me...? I did know the two numbers right away, once they told me the ratio $r$!

Question: What are these mysterious numbers $x$ and $y$?

Notes

  • This is the second puzzle that I created on my own, besides the "consecutive-ranks" puzzle.
  • I could resume the conversation in just two lines but I preferred to ameliorate it to ease the solution and make it more adapted to logic.
  • The solution is unique.
  • The uniqueness is twice verified :D
  • I will show the solution if this post becomes be sufficiently downvoted or if there are enough unsuccessful attempts
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  • 1
    $\begingroup$ Clarification: In the third statement, Mathematician 1 says that he knows that mathematician 2 knows that 1 doesn't know the answer. But then says that he (1) doesn't know if 2 doesn't know that 1 knows that 2 doesn't. In this context, that final "doesn't"; what is that referring to? Is that "doesn't know x and y from s"? Or is he still referring to 2 knowing what 1 knows? $\endgroup$ – Trevor Powell Feb 22 '15 at 4:09
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    $\begingroup$ @Abidare001: How does the ratio guy knowing the numbers imply that one divides the other? If the ratio was something like 79/73, the ratio guy would know the numbers immediately. $\endgroup$ – user2357112 Feb 22 '15 at 8:55
  • $\begingroup$ @user2357112 79/73=1 and 1 can be a quotient for many other couples $\endgroup$ – Abr001am Feb 22 '15 at 10:05
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    $\begingroup$ ...Since when is 79/73=1? Is this programming language floor division? $\endgroup$ – Lopsy Feb 22 '15 at 12:06
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    $\begingroup$ If it's floor or truncating division, this problem still doesn't make any sense. Those operations with numbers in this range can produce any integer from 1 to 50, and any integer from 1 to 49 can be produced at least two ways. For the ratio guy to know the numbers immediately, the ratio would have to be 50 and the numbers would have to be 100 and 2, but that contradicts the fact that the difference guy couldn't figure it out. This problem is simply inconsistent. $\endgroup$ – user2357112 Feb 22 '15 at 18:23
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Note:

The proof of impossibility presented in this answer is no longer valid; its logic was rendered invalid by additional information added later, in the fifth (!) revision to the puzzle definition. I'm leaving this answer here only for historical context; it is not a valid answer to the (current) revised form of the question.

For ease of reference, I state below the old (fourth) version of the puzzle to which my solution refers.


Statement of the fourth version of the puzzle

Three perfect mathematicians with extremely strong memories are taking an exam. The examiner tells each of them a certain piece of information about $x$ and $y$, which two positive integers between 2 and 100:

  • The first one is told the difference $d=x-y$
  • The second one is told the sum $s=x+y$
  • The third one the ratio $r=x/y$.

Then the following conversation takes place:

  • First mathematician: I cannot deduce $x$ and $y$ from their difference $d$.

  • Second mathematician: I knew that you cannot.

  • First mathematician: Don't bother to say that, I already knew that you knew that I cannot. But I'm not sure if you don't know that I know that you don't !!!!

  • Third mathematician: What about me...? I did know the two numbers right away, once they told me the ratio $r$!

It's self-evident from the last declaration that one of these two numbers divide the other.

Question: What are these mysterious numbers $x$ and $y$?


The solution of the fourth version of the puzzle

There is no solution.

Proof:

First mathematician:

The only pair of $(x,y)$ values which the first mathematician could identify without consulting the others, just based on the $d$ value, would be $(100,2)$, which would produce a $d$ value of $98$ (its maximum possible value). There is no other combination of $(x,y)$ values which could produce that $d$ value, and there is no other $d$ value which has a unique set of $(x,y$) values which could produce it. That the first mathematician doesn't know the values tells us that $(x,y)$ are not $(100,2)$, and $d$ is not $98$.

Second mathematician:

The only values that the second mathematician could identify without consulting the others, just based on the $s$ value, would be $(2,2)$ and $(100,100)$ ($s$ values of $4$ and $200$, respectively). Note that it's never explicitly stated that the second mathematician doesn't know the values, so $(2,2)$ and $(100,100)$ are still viable answers (so far).

What the second mathematician knows about the first one:

But the second mathematician does know that the first mathematician doesn't know the values. This means that based upon the $s$ value he was given, the second mathematician knows that the $d$ value cannot be $98$. Which means that the $s$ value received by the second mathematician cannot be $102$ (since that's the only $s$ value which could produce the $d=98$ value which would allow the first mathematician to immediately know $(x,y)$). Although there are many $(x,y)$ pairs which would produce $s=102$, if $s=102$ then the second mathematician can't be certain that the first mathematician didn't get a $d$ value of $98$, and therefore can't possibly know $(x,y)$.

What the first mathematician knows about what the second knows about the first:

The first mathematician states that he knows that the second mathematician is certain that the first mathematician doesn't know the answer. That is, from the $d$ value he received (which we know isn't $98$), the first mathematician must be able to deduce that the $s$ value the second mathematician received could not be $102$, because if the $s$ value had been $102$, then the second mathematician would reason that the $d$ value might have been $98$, which would mean that the first mathematician might know the $(x,y)$ values.

SO. (my head hurts)

From the first mathematician's $d$ value, he somehow knew that $s$ absolutely couldn't be $102$. So as an example of this, his $d$ value couldn't be $2$. Because with $d=2$, an $(x,y)=(52,50)$ would give $s=102$, which would not allow the second mathematician to exclude a possible $d$ value of $98$; With a $d$ value of $2$, the first mathematician can't be certain that the second mathematician didn't receive an $s$ value of $102$, and if the second mathematician could have received an $s$ value of $102$, he couldn't have said definitively that the first mathematician didn't know the answer. Similarly, a $d$ value of $96$ would admit a result of $(99, 3)$ which would have given $s=102$; same problem.

The only solution to 1-knows-2-knows-1-doesn't know:

As it turns out, there's only one way to be sure, based only on the $d$ value, that the $s$ value cannot be $102$ (and that therefore the second mathematician must know that the first mathematician absolutely cannot know the answer); and that's if the $d$ value received by the first mathematician was an odd number. If $d$ is odd, then just one of $x$ or $y$ must be odd, and that means that $s$ must also be odd (since $odd + even = odd$). So based on knowing an odd $d$ value, the first mathematician also knows that $s$ is odd, and therefore cannot be $102$, and therefore the second mathematician knows that $d$ cannot be $98$, and therefore also knows that the first mathematician can't deduce the $(x,y)$ values based on the $d$ value. As a side-note, since we know that the third mathematician knows the ratio $r$, and we've been told that $r$ is an integer value with no remainder, we also know that $x$ must be even, and $y$ must be odd. (Because if $x$ was odd and $y$ was even, the ratio would always have a fractional part, which we've been told is not the case). This side-note isn't important to the 1-knows-2-knows-1-doesn't problem, but it's an important point, later. So I'll just leave it here.

Now let's look at the third mathematician, who was able to tell the values of $(x,y)$ just from their ratio, $r$.

Now, mathematician 3 doesn't know anything we've deduced above. He claims to have known the $x$ and $y$ values immediately upon hearing the $r$ value, which means that the $r$ value must be between $50$ and $34$, inclusive; as those $r$ values can only be reached with a single pair of $x$ and $y$ values (and all of them require $y$ to be $2$). Any $r$ values below $34$ could be arrived at in multiple ways (For example, $33 = 66/2$ or $33 = 99/3$, for example, and with an $r$ value below $34$, mathematician 3 can't determine which set of values are the correct ones without information that he doesn't yet have, from the other mathematicians).

Pulling all these elements together to prove that my answer is correct:

According to statements 1, 2, and the first half of 3 (from the first two mathematicians), we can deduce that the $y$ value must be odd. According to statement 4 (from the third mathematician), we can deduce that the $y$ value must be $2$. Since there are no positive integers which are simultaneously $2$ and odd (citation needed), there cannot be a solution to the puzzle, as presented. If $y$ is not odd, then mathematician 1 cannot make the first half of statement 3. If $y$ is not $2$, then mathematician 3 cannot make statement 4.

Q.E.D.

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  • $\begingroup$ I took "But I'm not sure if you don't know that I know that you don't !!!!" specifically, "I know that you don't" to mean 2nd mathematician didn't know answer just from the sum. $\endgroup$ – Marmy1954 Feb 22 '15 at 6:24
  • $\begingroup$ The double-negatives in that bit of the puzzle really hurt my head. :) $\endgroup$ – Trevor Powell Feb 22 '15 at 6:38
  • $\begingroup$ @TrevorPowell if it did hurt ur head when trying to resolve it ........ guess what happened to my head tryin to create a unique solution :D $\endgroup$ – Abr001am Feb 22 '15 at 11:21
  • $\begingroup$ your analysis is right but let me tell you that you didnt even cross the first stage of solution , All the puzzle's difficulty resides in the third statement , try harder and i advice you to deal with it by coding . good luck , this effort s worth to be upvoted :) $\endgroup$ – Abr001am Feb 22 '15 at 14:03
  • $\begingroup$ I've converted the first part to LaTeX, but ran out of time; if someone wants to convert the rest, feel free, otherwise I may come back later and finish it. $\endgroup$ – Aza Feb 23 '15 at 1:30
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Partial answer:

Assuming that "one is divisible by the other" means that one is evenly divisible by the other so that there is no remainder, then merely the 3rd mathematician's statement narrows is down to:

x = even integer between 68 and 100 inclusive
y = 2
Because r has to be a whole number that can only be achieved with a unique set of x & y values. I confirmed that these are the only unique cases through simple brute force.

Adding in the first statement (as discussed in other comments), we can drop the case

x = 100, y = 2

I can't get wrap my head around narrowing it down past that. I've narrowed it down to just 16 of the 9,801 possibilities but I can't get any further. I'm left with:

x = {68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98}
y = 2

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  • $\begingroup$ each correct illustrated answer would be approved three days since last time modified .you have plenty of time to review your answer . Countdown fom now . $\endgroup$ – Abr001am Feb 23 '15 at 21:42
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The key here is to remember that we have that one number divides the other as common knowledge.

Statements number 1-4:

Statement 1:

There are quite a few $d$ values that are being avoided here: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 74, 82, 85, 86, 87, 91, 93, 94, 95, 98. All of these have only only way to be formed under the rule.

Statement 2:

The corresponding $s$ values that have to be avoided are: 6, 9, 15, 21, 33, 39, 51, 57, 69, 78, 86, 87, 90, 93, 93, 95, 98, 99, 102, 105, 105, 111, 123, 129, 141.

Statement 3a:

This gives a new set of avoided $d$ values: 9, 15, 26, 27, 30, 33, 34, 35, 45, 51, 52, 54, 57, 60, 63, 66, 68, 70, 72, 75, 77, 78, 80, 81, 84, 90, 96.

Statement 3b:

I think what this is saying is that the possible $s$ values for the actual $d$ value includes one that could lead to one of the above (3a excluded) $d$ values. These possible $d$ values are 10, 18, 20, 21, 22, 24, 25, 28, 32, 36, 39, 40, 42, 44, 48, 49, 50, 55, 56, 62, 64, 65, 69, 76, 88, 92.

Statement 4:

This means that the ratio must be over 33, and the only qualifying pair remaining for this is (2,90).

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  • $\begingroup$ each correct illustrated answer would be approved three days since last time modified .you have plenty of time to review your answer . Countdown fom now . $\endgroup$ – Abr001am Feb 23 '15 at 21:41
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Assuming you mean "between 2 and 100" to be inclusive. x=

98

y=

2

because

A higher ratio limits the (x,y) possibilities. The max ratio would r=50, where (x,y) = (100,2) but then mathematician 1 could tell answer because d=98 has only one solution. Next highest ratio is r=49, ==> (x,y)= (98,2). Since d=96 has 2 solutions, (98,2) (100,4), math 1 would not know answer. And sum=100 has multiple solutions so 2nd mathematician would also not know answer. Therefore, third mathematician was told that r=49.

How mathematician1 knows mathematician2 knows mathematician1 doesn't know answer. Or How "I know that you know that I dont know"

Since mathematician#1 would get d=96 ==> (x,y)=(98,2) or (100,4).

IF former,

That is for (x,y)=(98,2)then mathematician#2 would get s=100. S=100 has multiple solutions giving integer ratio. EX: (98,2), (96,4), (95,5),(80,20). And each of these solutions have differences that yield multiple solutions, so Mathematician #2 knows that mathematician#1 does not know, and mathematician #1 can reason that.

IF latter,

That is for (x,y)=(100,4), then math#2 gets s=104, which also has at least 2 solutions (100,4),(91,13). Each of these have differences that have multiple solutions. (100,4) => d=96, which has at least 2 solutions. (91,13) => d=78, (80,2), (81,3).

To find pairs

use x-y=d, where d is known, and x/y= integer => x=my for some m, so my-y=(m-1)y=d. So I factored the difference and let y be a factor and then added y to d to get x.

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  • $\begingroup$ In the puzzle, Mathematician 1 states (rather rudely) that he knows that Mathematician 2 knows that 1 doesn't know the values of (x,y). If all 1 knows is that d=96, how does he know for certain that 2 knows that 1 doesn't know the xy values? $\endgroup$ – Trevor Powell Feb 22 '15 at 6:34
  • $\begingroup$ Looks like double negatives got me too. Added to answer. I did this intuitively, going for the highest ratio and eliminating r=100/2 because there were not multiple solutions for d=98. $\endgroup$ – Marmy1954 Feb 22 '15 at 17:28
  • $\begingroup$ @Marmy1954 this solution u v proposed dosnt provoke this statement "I know that you know that I dont know" -good luck further ... btw ur nick seems to befamiliar to me :D $\endgroup$ – Abr001am Feb 22 '15 at 20:00
  • $\begingroup$ each correct illustrated answer would be approved three days since last time modified .you have plenty of time to review your answer . Countdown fom now . $\endgroup$ – Abr001am Feb 23 '15 at 21:43
  • $\begingroup$ I reworded my explanation under "if former". Hopefully it is clearer. The If former and If later explanations are meant to explain how #1 knows that #2 knows that #1 does not know. $\endgroup$ – Marmy1954 Feb 24 '15 at 1:41
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My rewording of the question; I think it is identical:

  1. First mathematician: I cannot deduce x and y from their difference d .
  2. Second mathematician: I knew that you cannot.
  3. First mathematician: Don't bother to say that, I already knew that you knew that I cannot.
  4. Second mathematician: Hm
  5. First mathematician: Well... you still don't know what the numbers are.
  6. Second mathematician: Ah but did I already know that you knew that? • First mathematician: I don't know...
  7. Third mathematician: Losers... I did know the two numbers right away, once they told me the ratio r !

This shows how I interpret the question which will be worked through below.


The above statement does not seem to have any possible answers. In order to solve this I created a simple workbook which determines all possible $s$ values if there is a given $d$ and vice versa. I then listed all possible values for $d$ and $s$ ignoring any of the statements. Note that if you know $s$ and $d$ you know the numbers $x$ and $y$. If you don't, you don't.

First I counted the number of potential $s$ values for a given $d$. If the number is $1$, anyone with that $d$ would know the numbers. There were $25$ such values of $d$ (called $list1$) and $63$ where he wouldn't know ($list2$). The remaining $12$ are impossible as there are $0$ corresponding values of $s$.

Second I counted the number of potential $d$ values for a given $s$ from $list1$ or $list2$. If the number from $list1=0$ but the number from $list2>0$ then statement $2$ is true for that value $s$. This $list3$ contains $110$ values of $s$.

Next came $list4$ which would only contain $d$ values which meet statement 3. This is the list of $d$ values where all corresponding values of $s$ are from $list3$. $list4$ has $35$ entries.

Line 4 of the above formulation of the puzzle is used to indicate that the second guy has been following all of this. $list5$ is generated as the list of all possible $s$ values from the $d$ values of $list4$. This should be indicated by the tense (past, present, future) of the original wording of the question. $list5$ has 62 entires. It should be noted that there are only 3 values from $list4$ which will work well with statement 7: $76$,$88$, and $92$.

Line 5 has two parts: First we generate $list6$ (whew) from $list5$ which only contains values of $s$ which could only result in $1$ corresponding values of $d$ from $list4$. This has $30$ values and represents the cases where guy 2 currently knows the numbers. Second we generate $list7$ of $d$ values which do not correspond with any $s$ values from $list6$. This is the list of $d$ values for which Line 5 is correct (guy 1 knows guy 2 doesn't know) and only has $11$ values including $76$ and $92$ but not $88$.

For ease of use, we generate $list8$ which contains all $d$ values which has both corresponding $s$ values which would cause guy 2 to know the numbers and corresponding $s$ values which would not cause guy 2 to know the numbers. This list contains values where guy 1 doesn't know that guy 2 doesn't know the numbers. (note: $d=58$ seems to be the only case where guy 1 knows that guy 2 does know the numbers.)

Here is where we run into an issue. $d=76$ means $s=80$,$84$, or $114$. $d=92$ means $s=96$ or $100$. All of these $s$ values correspond with $d$ values from both $list7$ and $list8$. This means anyone with that $s$ number despite all this would not know before line 5 that line 5 was true. The $d$ values that are still viable after line 6 (but not line 7) are: $8$, $12$, $14$, $16$, $20$, $24$, and $28$.

Please let me know whether you find a mistake, i misunderstood the question, and/or what you think the answer should be. Wording always seems to be the issue with your questions like this and if you think there is an answer, this is likely the issue.

I think if anything is misunderstood it is whether line 4 exists.

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  • $\begingroup$ Please note that I have attempted to solve this from a list of possible groups (x,y,r,d,s) instead of looking at possible d and s values. I recieved the same result. $\endgroup$ – kaine Feb 27 '15 at 20:02
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magic numbers are : (78 , 2 )

First off lets note a_b a couple of two numbers (a*b,b)

Starting from last statement , any set with absent couple such as (2,x) where x > 33 will be excluded

Such couples are:

ratio=35  2-35 
ratio=36  2-36 
ratio=37  2-37 
ratio=38  2-38 
ratio=39  2-39 
ratio=40  2-40 
ratio=41  2-41 
ratio=42  2-42 
ratio=43  2-43 
ratio=44  2-44 
ratio=45  2-45 
ratio=46  2-46 
ratio=47  2-47 
ratio=48  2-48 
ratio=49  2-49  
ratio=50  2-50 

So ... standing on that ground , lets begin to work .

this list contains sets of differences that subtractor doesnt know .

set of difference 68 :  2_35 = (70, 35)  4_18 = (72, 18)  17_5 = (85, 5)  

set of difference 70 :  2_36 = (72, 36)  5_15 = (75, 15)  7_11 = (77, 11)  10_8 = (80, 8)  14_6 = (84, 6)  

set of difference 72 :  2_37 = (74, 37)  3_25 = (75, 25)  4_19 = (76, 19)  6_13 = (78, 13)  8_10 = (80, 10)  9_9 = (81, 9)  12_7 = (84, 7)  18_5 = (90, 5)  24_4 = (96, 4)  

set of difference 76 :  2_39 = (78, 39)  4_20 = (80, 20)  19_5 = (95, 5)  

set of difference 78 :  2_40 = (80, 40)  3_27 = (81, 27)  6_14 = (84, 14)  13_7 = (91, 7)  

set of difference 80 :  2_41 = (82, 41)  4_21 = (84, 21)  5_17 = (85, 17)  8_11 = (88, 11)  10_9 = (90, 9)  16_6 = (96, 6)  20_5 = (100, 5)  

set of difference 84 :  2_43 = (86, 43)  3_29 = (87, 29)  4_22 = (88, 22)  6_15 = (90, 15)  7_13 = (91, 13)  12_8 = (96, 8)  14_7 = (98, 7)  

set of difference 88 :  2_45 = (90, 45)  4_23 = (92, 23)  8_12 = (96, 12)  11_9 = (99, 9)  

set of difference 90 :  2_46 = (92, 46)  3_31 = (93, 31)  5_19 = (95, 19)  6_16 = (96, 16)  9_11 = (99, 11)  10_10 = (100, 10)  

set of difference 92 :  2_47 = (94, 47)  4_24 = (96, 24)  

set of difference 96 :  2_49 = (98, 49)  3_33 = (99, 33)  4_25 = (100, 25)

this list contains elements grouped by their similar sum , accepted if summer knows for sure that subtractor doesnt know.

set of sum = 66  :
2_34  -temporarily accepted-  5_13  -temporarily accepted-  7_9  -temporarily accepted-  10_6  -temporarily accepted-  14_4  -temporarily accepted-  

this list is accepted 


set of sum = 68  :
2_35  -temporarily accepted-  3_23  -temporarily accepted-  4_17  -temporarily accepted-  6_11  -temporarily accepted-  8_8  -temporarily accepted-  9_7  -temporarily accepted-  12_5  -temporarily accepted-  18_3  -temporarily accepted-  24_2  -temporarily accepted-  

this list is accepted 


set of sum = 70  :
2_36  -temporarily accepted-  

this list is accepted 


set of sum = 72  :
2_37  -temporarily accepted-  4_18  -temporarily accepted-  19_3  -temporarily accepted-  

this list is accepted 


set of sum = 74  :
2_38  -rejected because 2_38 is unique in the other side of differences- 
 3_25  6_12  13_5  26_2  

this list is rejected 


set of sum = 76  :
2_39  -temporarily accepted-  4_19  -temporarily accepted-  5_15  -temporarily accepted-  8_9  -temporarily accepted-  10_7  -temporarily accepted-  16_4  -temporarily accepted-  20_3  -temporarily accepted-  

this list is accepted 


set of sum = 78  :
2_40  -temporarily accepted-  

this list is accepted 


set of sum = 80  :
2_41  -temporarily accepted-  3_27  -temporarily accepted-  4_20  -temporarily accepted-  6_13  -temporarily accepted-  7_11  -temporarily accepted-  12_6  -temporarily accepted-  14_5  -temporarily accepted-  21_3  -temporarily accepted-  28_2  -temporarily accepted-  

this list is accepted 


set of sum = 82  :
2_42  -rejected because 2_42 is unique in the other side of differences- 


this list is rejected 


set of sum = 84  :
2_43  -temporarily accepted-  4_21  -temporarily accepted-  8_10  -temporarily accepted-  11_7  -temporarily accepted-  22_3  -temporarily accepted-  

this list is accepted 


set of sum = 86  :
2_44  -rejected because 2_44 is unique in the other side of differences- 
 3_29  5_17  6_14  9_9  10_8  15_5  18_4  30_2  

this list is rejected 


set of sum = 88  :
2_45  -temporarily accepted-  4_22  -temporarily accepted-  23_3  -temporarily accepted-  

this list is accepted 


set of sum = 90  :
2_46  -temporarily accepted-  

this list is accepted 


set of sum = 92  :
2_47  -temporarily accepted-  3_31  -temporarily accepted-  4_23  -temporarily accepted-  6_15  -temporarily accepted-  8_11  -temporarily accepted-  12_7  -temporarily accepted-  16_5  -temporarily accepted-  24_3  -temporarily accepted-  32_2  -temporarily accepted-  

this list is accepted 


set of sum = 94  :
2_48  -rejected because 2_48 is unique in the other side of differences- 
 7_13  14_6  

this list is rejected 

the previous informations are unnecessary until this point: where subtractor declares his important information:

A list which fits third statement:

set of difference= 68  : 
 2_35  4_18       17_5       rejected because 

 it exists a couple (ab,b)=(85,5) in this set where 2_50 is unique in his set of difference where 
 2_50 has same sum of a_b = 17_5 

 the summer wudnt say i know that you know that i dont 



set of difference= 70  : 
 2_36  5_15  7_11  10_8       rejected because

 it exists a couple (ab,b)=(80,8) in this set where 2_44 is unique in his set of difference where 
 2_44 has same sum of a_b = 10_8 

 the summer wudnt say i know that you know that i dont 



set of difference= 72  : 
 2_37       3_25       rejected because 

 it exists a couple (ab,b)=(75,25) in this set where 2_38 is unique in his set of difference where 
 2_38 has same sum of a_b = 3_25 

 the summer wudnt say i know that you know that i dont 



set of difference= 76  : 
 2_39               4_20               19_5          
 this is the only set accepted because 

 2_39   :  2_39 has same sum of 
 5_15 which has same diff of 
 2_36 which is unique in his set of same sum  

 4_20   :  4_20 has same sum of 
 3_27 which has same diff of 
 2_40 which is unique in his set of same sum  

 19_5        accepted because there dosnt exist any such unique couple  

this set contains the unique solution because there exists two couple  2_39 , 19_5 
 where the subtractor is not sure whether the summer knows if the subtractor knows that summer dosent know the numbers



set of difference= 78  : 
 2_40  3_27  6_14       rejected because 

 it exists a couple (ab,b)=(84,14) in this set where 2_44 is unique in his set of difference where 
 2_44 has same sum of a_b = 6_14 

 the summer wudnt say i know that you know that i dont 



set of difference= 80  : 
 2_41  4_21       5_17       rejected because 

 it exists a couple (ab,b)=(85,17) in this set where 2_44 is unique in his set of difference where 
 2_44 has same sum of a_b = 5_17 

 the summer wudnt say i know that you know that i dont 



set of difference= 84  : 
 2_43       3_29       rejected because 

 it exists a couple (ab,b)=(87,29) in this set where 2_44 is unique in his set of difference where 
 2_44 has same sum of a_b = 3_29 

 the summer wudnt say i know that you know that i dont 



set of difference= 88  : 

 2_45   :  2_45 has same sum of 
     23_3 which has same diff of 
 46_2 which is unique in his set of same sum  

 4_23   :  4_23 has same sum of 
 3_31 which has same diff of 
 2_46 which is unique in his set of same sum  

 8_12   :  8_12 has same sum of 
 13_7 which has same diff of 
 2_40 which is unique in his set of same sum  

 11_9   :  11_9 has same sum of 
 10_10 which has same diff of 
 2_46 which is unique in his set of same sum  


 rejected because all the couples in this set leads to atleast one unique couple in the list of addition  
 which means subtractor knows cetainly that summer doenst know if subtractor knows wheather summer doesnt know numbers 



set of difference= 90  : 
 2_46  3_31  5_19  6_16       rejected because 

 it exists a couple (ab,b)=(96,16) in this set where 2_50 is unique in his set of difference where 
 2_50 has same sum of a_b = 6_16 

 the summer wudnt say i know that you know that i dont 



set of difference= 92  : 

 2_47   :  2_47 has same sum of 
 3_31 which has same diff of 
 2_46 which is unique in his set of same sum  

 4_24   :  4_24 has same sum of 
 5_19 which has same diff of 
 2_46 which is unique in his set of same sum  


 rejected because all the couples in this set leads to atleast one unique couple in the list of addition  
 which means subtractor knows cetainly that summer doenst know if subtractor knows wheather summer doesnt know numbers 



set of difference= 96  : 
 2_49  3_33       rejected because 

 it exists a couple (ab,b)=(99,33) in this set where 2_50 is unique in his set of difference where 
 2_50 has same sum of a_b = 3_33 

 the summer wudnt say i know that you know that i dont 

from the last list there s just one accepted set , where there s just one accepted couple specified from last divisor's declaration.


$\endgroup$
  • $\begingroup$ no choice left after 8 days delay $\endgroup$ – Abr001am Mar 1 '15 at 19:10
  • $\begingroup$ The reason this answer was not found by anyone is because both "5_15" and "3_27" were eliminated in the first part of line 3. With this being the case, subtractor would know that summer didn't know that d knew that s didn't know. This answer might make sense if summer was not told the first part of line 3. A question that would fit with this answer would replace the second half of line 3 with "I don't know if you knew (past tense) that I knew that you didn't." I have not confirmed this doesn't cause any other answers to become correct. $\endgroup$ – kaine Mar 2 '15 at 15:01
  • $\begingroup$ trying to phrase this delicately Am I correct that according to your interpretation Mathematician 1 would assume that Mathematician 2 could not use the first part of line 3 when considering the second part of line 3? $\endgroup$ – kaine Mar 2 '15 at 15:12
  • $\begingroup$ @kaine the sets containing "5_15" and "3_27" are excluded refering to first part of line 3 , final set is fruited from an intersection between two resulted sets of first and second part of line 3 . $\endgroup$ – Abr001am Mar 5 '15 at 10:55

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