Picking marbles











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We have 15 urns each of them having a different number of marbles, from 1 to 15. We start by picking the same number of marbles from each of the urns we choose. We repeat the process until we have picked all marbles. What is the minimum number of days we can finish picking all marbles? Just to clarify that it is not necessary to pick marbles from EVERY urn.




I don't think I can make it in less than 5 moves (start by picking 6, then 4, then 3, then 2 and 1) but I am fairly sure it can be done in 4 or maybe less.



Any ideas?










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  • What do you mean by picking 6?
    – Akash Roy
    Nov 27 at 14:17















up vote
7
down vote

favorite
2













We have 15 urns each of them having a different number of marbles, from 1 to 15. We start by picking the same number of marbles from each of the urns we choose. We repeat the process until we have picked all marbles. What is the minimum number of days we can finish picking all marbles? Just to clarify that it is not necessary to pick marbles from EVERY urn.




I don't think I can make it in less than 5 moves (start by picking 6, then 4, then 3, then 2 and 1) but I am fairly sure it can be done in 4 or maybe less.



Any ideas?










share|cite|improve this question
























  • What do you mean by picking 6?
    – Akash Roy
    Nov 27 at 14:17













up vote
7
down vote

favorite
2









up vote
7
down vote

favorite
2






2






We have 15 urns each of them having a different number of marbles, from 1 to 15. We start by picking the same number of marbles from each of the urns we choose. We repeat the process until we have picked all marbles. What is the minimum number of days we can finish picking all marbles? Just to clarify that it is not necessary to pick marbles from EVERY urn.




I don't think I can make it in less than 5 moves (start by picking 6, then 4, then 3, then 2 and 1) but I am fairly sure it can be done in 4 or maybe less.



Any ideas?










share|cite|improve this question
















We have 15 urns each of them having a different number of marbles, from 1 to 15. We start by picking the same number of marbles from each of the urns we choose. We repeat the process until we have picked all marbles. What is the minimum number of days we can finish picking all marbles? Just to clarify that it is not necessary to pick marbles from EVERY urn.




I don't think I can make it in less than 5 moves (start by picking 6, then 4, then 3, then 2 and 1) but I am fairly sure it can be done in 4 or maybe less.



Any ideas?







combinatorics






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edited Nov 27 at 14:30

























asked Nov 27 at 14:12









Reyansh Laghari

1515




1515












  • What do you mean by picking 6?
    – Akash Roy
    Nov 27 at 14:17


















  • What do you mean by picking 6?
    – Akash Roy
    Nov 27 at 14:17
















What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17




What do you mean by picking 6?
– Akash Roy
Nov 27 at 14:17










2 Answers
2






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oldest

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up vote
4
down vote



accepted










You can look at your urns as an array of 4 bit integers:
$0001_b$
$0010_b$
$0011_b$

...
$1111_b$



On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.



In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.






share|cite|improve this answer






























    up vote
    3
    down vote













    It is possible in 4 days:



    First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.



    Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.



    Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.



    Last day you took balls from all the nonemty urns.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      4
      down vote



      accepted










      You can look at your urns as an array of 4 bit integers:
      $0001_b$
      $0010_b$
      $0011_b$

      ...
      $1111_b$



      On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.



      In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.






      share|cite|improve this answer



























        up vote
        4
        down vote



        accepted










        You can look at your urns as an array of 4 bit integers:
        $0001_b$
        $0010_b$
        $0011_b$

        ...
        $1111_b$



        On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.



        In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.






        share|cite|improve this answer

























          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          You can look at your urns as an array of 4 bit integers:
          $0001_b$
          $0010_b$
          $0011_b$

          ...
          $1111_b$



          On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.



          In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.






          share|cite|improve this answer














          You can look at your urns as an array of 4 bit integers:
          $0001_b$
          $0010_b$
          $0011_b$

          ...
          $1111_b$



          On every step you can set one bit to $0$ on every integer for which it isn't already 0. There are 4 bits so you can do it in 4 steps. If we go back to decimal, you're removing 8, then 4, then 2, then 1.



          In fact we can also prove that $n$ is the minimum number of steps for $n$-digit urns through a recursion on the number of digits.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 27 at 15:24

























          answered Nov 27 at 15:19









          Rchn

          49015




          49015






















              up vote
              3
              down vote













              It is possible in 4 days:



              First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.



              Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.



              Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.



              Last day you took balls from all the nonemty urns.






              share|cite|improve this answer

























                up vote
                3
                down vote













                It is possible in 4 days:



                First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.



                Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.



                Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.



                Last day you took balls from all the nonemty urns.






                share|cite|improve this answer























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  It is possible in 4 days:



                  First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.



                  Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.



                  Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.



                  Last day you took balls from all the nonemty urns.






                  share|cite|improve this answer












                  It is possible in 4 days:



                  First day you reduce the number of balls by 8 in urns with at least 8 balls. So now each urn has at most 7 balls.



                  Second day you reduce the number of balls by 4 in urns with at least 4 balls. So now each urn has at most 3 balls.



                  Third day you reduce the number of balls by 2 in urns with at least 2 balls. So now each urn has at most 1 ball.



                  Last day you took balls from all the nonemty urns.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 at 15:19









                  greedoid

                  35.2k114489




                  35.2k114489






























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