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 1 hour ago









      Max0815Max0815

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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
            1 hour ago










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










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










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










          • $begingroup$
            @Max0815 You're welcome! Fun puzzle, thanks! :)
            $endgroup$
            – Bram28
            1 hour 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 1 hour ago

























          answered 1 hour ago









          Bram28Bram28

          61.7k44793




          61.7k44793












          • $begingroup$
            Thanks! I got it!
            $endgroup$
            – Max0815
            1 hour ago










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


















          • $begingroup$
            Thanks! I got it!
            $endgroup$
            – Max0815
            1 hour ago










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
















          $begingroup$
          Thanks! I got it!
          $endgroup$
          – Max0815
          1 hour ago




          $begingroup$
          Thanks! I got it!
          $endgroup$
          – Max0815
          1 hour ago












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




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










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










          • $begingroup$
            Yes, that's correct. Sorry for the error.
            $endgroup$
            – Robert Shore
            1 hour 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 59 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 1 hour 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
            1 hour ago










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










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












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




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


















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