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The two perfect logicians $A$ and $C$ meet their colleague $B$ at an exam, and they are not aware of the fact that $B$ is a moron and a horrible mathematician. At the exam, the examiner informs them that he has chosen two different integers $x$ and $y$ with $2\le x<y\le100$, such that $y$ is a multiple of $x$.

The examiner then tells the difference $d=y-x$ to the first mathematician $A$, the ratio $r=y/x$ to the second mathematician $B$, and the sum $s=x+y$ to the third mathematician $C$ in such a way that none knows which numbers have been whispered to the others.

The three mathematicians then start this implying conversation:

  • $A$: I don't know the numbers, and I know that you both know this.
  • $B$: I already knew the numbers when the examiner told me their ratio.
  • $A$ and $C$ simultaneously: Aha! We have just deduced the two numbers.
  • $B$: Oops, damn, I am sorry! I think I made a mistake in my calculation. As a matter of fact, the ratio $r$ does not allow me to deduce the two numbers.

Question1: What are those two numbers $x$ and $y$??

Question2: What are those two numbers $x$ and $y$, if the following conversation took place instead from the third line on??

  • $C$: Aha! I have just deduced the two numbers.
  • $A$: I strongly recommend for $B$ to recalculate his ratio because he was surely wrong according to my data.

For hints and notes, refer to the following links:
Deducing Two Numbers based on their Difference and Ratio
What are the numbers?
https://puzzling.stackexchange.com/questions/9525/what-are-the-numbers-third-edition

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  • 6
    $\begingroup$ All your linked previous puzzles were found to be broken. Please stop posting these. $\endgroup$ – xnor Feb 27 '15 at 21:54
  • $\begingroup$ Make a puzzle that has less options, such as 25X25. $\endgroup$ – Moti Feb 28 '15 at 17:21
  • $\begingroup$ @xnor the solution will be added like anytime before , just have patience . $\endgroup$ – Abr001am Feb 28 '15 at 18:57
  • $\begingroup$ @Abidare001 I am not convinced you should necessarily stop posting these but as I have found errors in 2 of your final solutions after having put more effort into finding answers to your questions than I have any other questions on this site ever looking (whew) you need to seriously put more effort into confirming your solutions before posting. That being said, i've not had enough time to consider this one yet. $\endgroup$ – kaine Mar 2 '15 at 15:20
  • $\begingroup$ Ok I might have been wrong to defend this. There doesn't seems to be a correct answer here either. $\endgroup$ – kaine Mar 2 '15 at 15:51
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Solution:

Magic numbers are :

4,23 = 92 and 4

first off lets note that such a coupe i,j is a representation of two numbers ij and i

Now , the first sentence in the intersection of two lists , the list of addition and division guy conditions , it is the following

5,12                11,6 

3,24                23,4        

2,47                4,24               

2,45                4,23               8,12               11,9               

2,33                4,17               8,9               16,5               32,3               

2,29                4,15               7,9               8,8               14,5               28,3               

you can verify the evidence of this list by this little xls tool

Xls file

this scripted excel sheet extracts the list of numbers which fits that the diference guy knows that summer knows that he doesnt know for sure , the second sheet shows how subtractor is sure about divisor knows that subtractor dosent know with a perfect explanation.

the final list is an intersection of those two lists.

now divisor knows the numbers which means j>33.

now both guys knows also which means the list will be truncated to :

2,47                4,24               

2,45                4,23               8,12               11,9               

the list concerning summer and intersecting the upper lists is

2,47        3,31        4,23        6,15         8,11        12,7       16,5         24,3         32,2  

2,45        4,22        23,3  

know divisor made a mistake which means such a couple 2,x is excluded , the only couple which appears twice in last two lists apart 2,x is 4,23

to be honest i cant go with explanation further than this.

thanks for your downvotes.


second Question:

the number is an intersection between this list

5,12                11,6 

3,24                23,4        

2,33                4,17               8,9               16,5               32,3               

2,29                4,15               7,9               8,8               14,5               28,3  

and this one

2,47        3,31        4,23        6,15         8,11        12,7       16,5         24,3         32,2  

2,45        4,22        23,3  

the only couple appeaering twice is 16,5

Which means numers in this case are

16 and 80

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  • $\begingroup$ 1st: Assume that 4,23 means x=23, y=92, r=y/x=4, d=y-x=69, and s=y+x=115. 2nd: Line 2 seems to read "I knew x and y (or i and j) immediately when r (or j) was told to me" this is inferred from your solution. If person A knows d=69, he immediately knows r=4 or 24. This means that he immediately knows that person B is wrong when he says he knew x and y immediately from r as that would require r>33. As A could not get believable information from a sentence he knows is incorrect; he could not possibly learn the numbers from B's statement. This answer is, therefore, provably wrong again. $\endgroup$ – kaine Mar 4 '15 at 15:03
  • $\begingroup$ Please let me know if this issue is my not understanding your english or what your data represents. The problem is clearly not that these problems are hard but either a miscommunication of the problem or (more likely) a flaw in your method of solving the problems. $\endgroup$ – kaine Mar 4 '15 at 15:08
  • $\begingroup$ "As A could not get believable information from a sentence he knows is incorrect; he could not possibly learn the numbers from B's statement. This answer is, therefore, provably wrong again" ... yes i agree , but the numbers which were in their minds , 2,45 for the subtractor and 2,47 for summer , were temporarily accepted until proven errounous by the last statement , the only choice left is 4,23 . $\endgroup$ – Abr001am Mar 4 '15 at 18:59
  • $\begingroup$ @kaine the answer was been edited there was something wrong at the beginning $\endgroup$ – Abr001am Mar 4 '15 at 19:04
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    $\begingroup$ Note that the "something wrong" being the actual ANSWER is very infuriating. Nevertheless, this one does actually seem to work. Good job. $\endgroup$ – kaine Mar 4 '15 at 21:27

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