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Two mathematicians are passing an exam, where the examiner informs them that he has chosen two different numbers between 2 and 100, such that one is divisible by the other. The examiner then tells the sum to the first mathematician A and the difference to the second mathematician B, in such a way that none knows which number has been whispered to other.

The two mathematicians start this discussion:

  • A: I don't know the numbers.
  • B: I already knew this.
  • A: I already knew that you knew this.
  • B: I already knew that you knew that I knew that you didn't know.
  • A: I still can't figure out what the two numbers are.
  • B: Oops, I think I miscalculated my last statement!!! As a matter of fact, I didn't know this already!!!

Question: What are these two numbers again??

For hints and notes, refer to the following links:
Deducing Two Numbers based on their Difference and Ratio
What are the numbers?

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    $\begingroup$ Is it $2\le x\lt y\le 100$ or $2\lt x\lt y\lt 100$? $\endgroup$ – Ivo Beckers Mar 26 '15 at 12:24
  • $\begingroup$ B never states that she doesn't know the numbers, only that she knew and then didn't know something about A's knowledge of the numbers. Can B state whether she knows the numbers, or is this intentionally part of the puzzle, and we are not to assume she does or does not know the numbers based on the information she had prior to discussing it with A? $\endgroup$ – Adam Davis Mar 26 '15 at 12:31
  • $\begingroup$ @IvoBeckers including 2 , this s a similar puzzle puzzling.stackexchange.com/questions/9499/… $\endgroup$ – Abr001am Mar 26 '15 at 12:50
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    $\begingroup$ @AshutoshNigam second statement does help. In 2 ways even. It helps because it eliminates possible solutions that have the same difference as the solutions eliminated by the first statement (otherwise B could not have know that A didn't know). Plus it eliminates all solutions that have a unique difference because otherwise B would know the solution at that point $\endgroup$ – Ivo Beckers Mar 26 '15 at 14:51
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    $\begingroup$ @Agawa001 Sorry I didn't answer this question before the bounty but I didn't see it. Very good job writing this one. I'm sorry others downvoted. $\endgroup$ – kaine Mar 28 '15 at 16:53
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Ok: There are 282 pairs of numbers $x,y$ such the $y/x$ is an integer greater than 1. The sum and difference are designated $s$ and $d$ respectively.

  1. If A doesn't know the numbers, any pairing which results in a unique value of $s$ should be eliminated. There are 34 such pairs.
  2. If B knew this, the value of $d$ must not have one of the 34 which would be eliminated in this way. Eliminate any pair with a value for $d$ which appears in the eliminated pairs. There are 75 eliminated in this step.
  3. If A knew this, the value of $s$ must not have one of the 109 which have been eliminated so far. Eliminate any pair with a value for $s$ which appears in the eliminated pairs (especially the 75 new ones). There are 113 eliminated in this step.
  4. If B were correct here, the value of $d$ must not have one of the 222 which have been eliminated so far. Eliminate any pair with a value for $d$ which appears in the eliminated pairs (especially the 113 new ones). This creates a list of 43 which A has been told are eliminated in this step and another list 17 which he has been told were not eliminated.
  5. From A's perspective, there are 17 pairs from which he can select the correct answer using his knowledge of $s$. As he cannot, $s$ must not be unique. There are 4 values for $s$ which are not unique from those 17 pairs: 15, 87, 93, and 99. These are the only currently possible values for $s$.
  6. The last line seems to mean that the correct pair was eliminated in step 4. Of the values eliminated in this step, only one has a value of $s$ which is one of the ones determined in step 5. The solution is therefore:

$x=33$, $y=66$, $s=99$, $d=33$

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  • $\begingroup$ oh man !!!!! whitout you and gamow there s no hope for these puzzles to be solved , im stenned of promptness of solution atleast let the puzzle breath :D - 50pts + upvote + right answer- $\endgroup$ – Abr001am Mar 28 '15 at 17:19
  • $\begingroup$ @Agawa001 I want to give you the bounty back. I've added a question like this I wrote a little while ago but didn't post because I thought it was too hard. I am going to apply the bounty from this on that (even it is means a reward for a good answer). $\endgroup$ – kaine Mar 30 '15 at 21:07

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