Logic problem on sum of possible numbers a given person can have if they had a conversation with another.












2












$begingroup$


Larry tells Marry and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Marry one of the numbers and then tells Jerry the other number. Then occurs a conversation between Marry and Jerry:



Marry: I don't know your number.



Jerry: I don't know your number either.



Marry: Ah, I now know your number.



Assuming both of them use correct logic, what is the sum of all possible numbers Marry could have?





What I have tried:




Marry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10



Jerry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10



Since Marry doesn't know Jerry's number, Marry's number could not have been 1 or 10.



Jerry's number then could not have been either 2 or 9 because then since he already knows Marry's # is not 1, then Marry's number would have been three. Same logic for 9.



So I am left with the possibilities as follows:



Marry's #s: 2, 3, 4, 5, 6, 7, 8, 9



Jerry's #s: 1, 3, 4, 5, 6, 7, 8, 10




I'm stuck here!



Help would be appreciated!



Also, it would also be nice if you would help me on this question(Transferring bases of numbers.) too!



Thanks!



Max0815










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$endgroup$

















    2












    $begingroup$


    Larry tells Marry and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Marry one of the numbers and then tells Jerry the other number. Then occurs a conversation between Marry and Jerry:



    Marry: I don't know your number.



    Jerry: I don't know your number either.



    Marry: Ah, I now know your number.



    Assuming both of them use correct logic, what is the sum of all possible numbers Marry could have?





    What I have tried:




    Marry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10



    Jerry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10



    Since Marry doesn't know Jerry's number, Marry's number could not have been 1 or 10.



    Jerry's number then could not have been either 2 or 9 because then since he already knows Marry's # is not 1, then Marry's number would have been three. Same logic for 9.



    So I am left with the possibilities as follows:



    Marry's #s: 2, 3, 4, 5, 6, 7, 8, 9



    Jerry's #s: 1, 3, 4, 5, 6, 7, 8, 10




    I'm stuck here!



    Help would be appreciated!



    Also, it would also be nice if you would help me on this question(Transferring bases of numbers.) too!



    Thanks!



    Max0815










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Larry tells Marry and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Marry one of the numbers and then tells Jerry the other number. Then occurs a conversation between Marry and Jerry:



      Marry: I don't know your number.



      Jerry: I don't know your number either.



      Marry: Ah, I now know your number.



      Assuming both of them use correct logic, what is the sum of all possible numbers Marry could have?





      What I have tried:




      Marry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10



      Jerry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10



      Since Marry doesn't know Jerry's number, Marry's number could not have been 1 or 10.



      Jerry's number then could not have been either 2 or 9 because then since he already knows Marry's # is not 1, then Marry's number would have been three. Same logic for 9.



      So I am left with the possibilities as follows:



      Marry's #s: 2, 3, 4, 5, 6, 7, 8, 9



      Jerry's #s: 1, 3, 4, 5, 6, 7, 8, 10




      I'm stuck here!



      Help would be appreciated!



      Also, it would also be nice if you would help me on this question(Transferring bases of numbers.) too!



      Thanks!



      Max0815










      share|cite|improve this question









      $endgroup$




      Larry tells Marry and Jerry that he is thinking of two consecutive integers from 1 to 10. He tells Marry one of the numbers and then tells Jerry the other number. Then occurs a conversation between Marry and Jerry:



      Marry: I don't know your number.



      Jerry: I don't know your number either.



      Marry: Ah, I now know your number.



      Assuming both of them use correct logic, what is the sum of all possible numbers Marry could have?





      What I have tried:




      Marry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10



      Jerry's #s: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10



      Since Marry doesn't know Jerry's number, Marry's number could not have been 1 or 10.



      Jerry's number then could not have been either 2 or 9 because then since he already knows Marry's # is not 1, then Marry's number would have been three. Same logic for 9.



      So I am left with the possibilities as follows:



      Marry's #s: 2, 3, 4, 5, 6, 7, 8, 9



      Jerry's #s: 1, 3, 4, 5, 6, 7, 8, 10




      I'm stuck here!



      Help would be appreciated!



      Also, it would also be nice if you would help me on this question(Transferring bases of numbers.) too!



      Thanks!



      Max0815







      logic






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      asked 50 mins ago









      Max0815Max0815

      49816




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          2 Answers
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          $begingroup$

          Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.



          So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).



          Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).



          Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.



          So, Mary's number is 2,3,8, or 9. Sum is 22






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I got it!
            $endgroup$
            – Max0815
            29 mins ago










          • $begingroup$
            @Max0815 You're welcome! Fun puzzle, thanks! :)
            $endgroup$
            – Bram28
            27 mins ago



















          1












          $begingroup$

          Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.



          If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.






          share|cite|improve this answer










          New contributor




          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













          • $begingroup$
            If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
            $endgroup$
            – Max0815
            30 mins ago










          • $begingroup$
            Yes, that's correct. Sorry for the error.
            $endgroup$
            – Robert Shore
            7 mins ago











          Your Answer





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          2 Answers
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          2 Answers
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          $begingroup$

          Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.



          So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).



          Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).



          Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.



          So, Mary's number is 2,3,8, or 9. Sum is 22






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I got it!
            $endgroup$
            – Max0815
            29 mins ago










          • $begingroup$
            @Max0815 You're welcome! Fun puzzle, thanks! :)
            $endgroup$
            – Bram28
            27 mins ago
















          2












          $begingroup$

          Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.



          So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).



          Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).



          Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.



          So, Mary's number is 2,3,8, or 9. Sum is 22






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I got it!
            $endgroup$
            – Max0815
            29 mins ago










          • $begingroup$
            @Max0815 You're welcome! Fun puzzle, thanks! :)
            $endgroup$
            – Bram28
            27 mins ago














          2












          2








          2





          $begingroup$

          Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.



          So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).



          Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).



          Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.



          So, Mary's number is 2,3,8, or 9. Sum is 22






          share|cite|improve this answer











          $endgroup$



          Note that the same reasoning applies to Jerry that you applied to Marry: Jerry's number cannot be 1 or 10, or otherwise Jerry would know Marry's number.



          So: Marry could indeed have 2 (or 9): Marry would initially indeed not know Jerry's number (since it would be 1 or 3 ... (or 8 or 10), but then since Jerry says he does not know Marry's (which makes sense if Jerry's is 3, for then for all Jerry knows Marry's is 2 or 4 and indeed in either case Mary would not know) Marry knows Jerry's cannot be 1 (or 10), and thus is 3 (or 8).



          Marry could also have 3 (8): Marry knows Jerry has 2 or 4 (7 or 9), but if Jerry has 2, then Jerry would know Marry must have 3 (for if Marry had 1 she would know Jerry has 2). So, Jerry saying he does not, she knows Jerry has 4 (7).



          Finally, Marry cannot have 4 (7): Jerry would have 3 or 5, but either way Jerry would not know, so Marry learns nothing from Jerry saying he does not know. Mary having 5 (6) also leaves too many options open for her to know Jerry's number on her second turn.



          So, Mary's number is 2,3,8, or 9. Sum is 22







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 28 mins ago

























          answered 32 mins ago









          Bram28Bram28

          61.7k44793




          61.7k44793












          • $begingroup$
            Thanks! I got it!
            $endgroup$
            – Max0815
            29 mins ago










          • $begingroup$
            @Max0815 You're welcome! Fun puzzle, thanks! :)
            $endgroup$
            – Bram28
            27 mins ago


















          • $begingroup$
            Thanks! I got it!
            $endgroup$
            – Max0815
            29 mins ago










          • $begingroup$
            @Max0815 You're welcome! Fun puzzle, thanks! :)
            $endgroup$
            – Bram28
            27 mins ago
















          $begingroup$
          Thanks! I got it!
          $endgroup$
          – Max0815
          29 mins ago




          $begingroup$
          Thanks! I got it!
          $endgroup$
          – Max0815
          29 mins ago












          $begingroup$
          @Max0815 You're welcome! Fun puzzle, thanks! :)
          $endgroup$
          – Bram28
          27 mins ago




          $begingroup$
          @Max0815 You're welcome! Fun puzzle, thanks! :)
          $endgroup$
          – Bram28
          27 mins ago











          1












          $begingroup$

          Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.



          If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.






          share|cite|improve this answer










          New contributor




          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













          • $begingroup$
            If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
            $endgroup$
            – Max0815
            30 mins ago










          • $begingroup$
            Yes, that's correct. Sorry for the error.
            $endgroup$
            – Robert Shore
            7 mins ago
















          1












          $begingroup$

          Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.



          If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.






          share|cite|improve this answer










          New contributor




          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













          • $begingroup$
            If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
            $endgroup$
            – Max0815
            30 mins ago










          • $begingroup$
            Yes, that's correct. Sorry for the error.
            $endgroup$
            – Robert Shore
            7 mins ago














          1












          1








          1





          $begingroup$

          Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.



          If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.






          share|cite|improve this answer










          New contributor




          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$



          Marry also knows that the numbers are consecutive. Therefore, if her number is $n$, she knows that Jerry's number is either $n+1$ or $n-1$. And after the fact that Jerry still doesn't know her number narrows down the field still further. Jerry's number can't be 1 or 10 because then he'd know Marry's number without help. Jerry's number also can't be 2 or 9 because then the fact that Marry doesn't know Jerry's number would tell Jerry that Marry's number has to be 3 or 8, respectively.



          If Jerry's inability to pinpoint Marry's number allows Marry to pinpoint Jerry's number, then her ability to eliminate 2 or 9, or the knowledge that Jerry's number isn't 1 or 10, must be new information that allows her to pinpoint her own number. Marry's number therefore has to be 2, 3, 8, or 9 and Jerry's number has to be 3, 4, 7, or 8.







          share|cite|improve this answer










          New contributor




          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer








          edited 5 mins ago





















          New contributor




          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 36 mins ago









          Robert ShoreRobert Shore

          715




          715




          New contributor




          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Robert Shore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.












          • $begingroup$
            If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
            $endgroup$
            – Max0815
            30 mins ago










          • $begingroup$
            Yes, that's correct. Sorry for the error.
            $endgroup$
            – Robert Shore
            7 mins ago


















          • $begingroup$
            If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
            $endgroup$
            – Max0815
            30 mins ago










          • $begingroup$
            Yes, that's correct. Sorry for the error.
            $endgroup$
            – Robert Shore
            7 mins ago
















          $begingroup$
          If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
          $endgroup$
          – Max0815
          30 mins ago




          $begingroup$
          If Marry's number is 2 or 9, couldn't she also be able to tell Jerry's number?(i.e. 3 or 9)
          $endgroup$
          – Max0815
          30 mins ago












          $begingroup$
          Yes, that's correct. Sorry for the error.
          $endgroup$
          – Robert Shore
          7 mins ago




          $begingroup$
          Yes, that's correct. Sorry for the error.
          $endgroup$
          – Robert Shore
          7 mins ago


















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