Picking marbles












7












$begingroup$



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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  • $begingroup$
    What do you mean by picking 6?
    $endgroup$
    – Akash Roy
    Nov 27 '18 at 14:17
















7












$begingroup$



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







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean by picking 6?
    $endgroup$
    – Akash Roy
    Nov 27 '18 at 14:17














7












7








7


2



$begingroup$



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







share|cite|improve this question











$endgroup$





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




combinatorics






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share|cite|improve this question













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share|cite|improve this question








edited Nov 27 '18 at 14:30







Reyansh Laghari

















asked Nov 27 '18 at 14:12









Reyansh LaghariReyansh Laghari

1666




1666












  • $begingroup$
    What do you mean by picking 6?
    $endgroup$
    – Akash Roy
    Nov 27 '18 at 14:17


















  • $begingroup$
    What do you mean by picking 6?
    $endgroup$
    – Akash Roy
    Nov 27 '18 at 14:17
















$begingroup$
What do you mean by picking 6?
$endgroup$
– Akash Roy
Nov 27 '18 at 14:17




$begingroup$
What do you mean by picking 6?
$endgroup$
– Akash Roy
Nov 27 '18 at 14:17










2 Answers
2






active

oldest

votes


















4












$begingroup$

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











$endgroup$





















    3












    $begingroup$

    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









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      4












      $begingroup$

      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











      $endgroup$


















        4












        $begingroup$

        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











        $endgroup$
















          4












          4








          4





          $begingroup$

          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











          $endgroup$



          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 '18 at 15:24

























          answered Nov 27 '18 at 15:19









          RchnRchn

          49015




          49015























              3












              $begingroup$

              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









              $endgroup$


















                3












                $begingroup$

                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









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  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









                  $endgroup$



                  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 '18 at 15:19









                  greedoidgreedoid

                  46k1160117




                  46k1160117






























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