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The grades of the students in a class


Past, Present and FutureWhat's the teacher's fractional addition algorithm?Students in a class with the same nameThe MathemagicianThe mysterious self-describing number #2Number Theory Class v2What a weird final examWho joined when?Cupcake divisionMy Graph Theory Students






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








3












$begingroup$


You are a new Math 102 professor in a university class of 88 students and would like to learn their grades from previous semester to know your students better and put a note of it to get an idea how good they are because the electronic system is broken right now.



But you need the precise grades from the last semester class Math 101, which could be integers between 0 to 100. You want to ask the least number of questions not to bother your students somehow or to show off, who knows... But you can ask only one kind of question:



"How many people got x,y,z... points from the last semester math class, raise your hand please?"



and take notes of the students accordingly whoever lift his/her hands. x,y,z,... could be any number of points/grades. such as;



"How many people got 0,15,30,32,38 points from the last semester math class, raise your hand please?"



So




What is the least number of questions you can ask to guarantee to know every single student grades in the class?











share|improve this question











$endgroup$









  • 2




    $begingroup$
    I assume that "you can ask only one question" means "you can ask only one kind of question"?
    $endgroup$
    – Gareth McCaughan
    9 hours ago










  • $begingroup$
    @GarethMcCaughan fixed.
    $endgroup$
    – Oray
    8 hours ago

















3












$begingroup$


You are a new Math 102 professor in a university class of 88 students and would like to learn their grades from previous semester to know your students better and put a note of it to get an idea how good they are because the electronic system is broken right now.



But you need the precise grades from the last semester class Math 101, which could be integers between 0 to 100. You want to ask the least number of questions not to bother your students somehow or to show off, who knows... But you can ask only one kind of question:



"How many people got x,y,z... points from the last semester math class, raise your hand please?"



and take notes of the students accordingly whoever lift his/her hands. x,y,z,... could be any number of points/grades. such as;



"How many people got 0,15,30,32,38 points from the last semester math class, raise your hand please?"



So




What is the least number of questions you can ask to guarantee to know every single student grades in the class?











share|improve this question











$endgroup$









  • 2




    $begingroup$
    I assume that "you can ask only one question" means "you can ask only one kind of question"?
    $endgroup$
    – Gareth McCaughan
    9 hours ago










  • $begingroup$
    @GarethMcCaughan fixed.
    $endgroup$
    – Oray
    8 hours ago













3












3








3





$begingroup$


You are a new Math 102 professor in a university class of 88 students and would like to learn their grades from previous semester to know your students better and put a note of it to get an idea how good they are because the electronic system is broken right now.



But you need the precise grades from the last semester class Math 101, which could be integers between 0 to 100. You want to ask the least number of questions not to bother your students somehow or to show off, who knows... But you can ask only one kind of question:



"How many people got x,y,z... points from the last semester math class, raise your hand please?"



and take notes of the students accordingly whoever lift his/her hands. x,y,z,... could be any number of points/grades. such as;



"How many people got 0,15,30,32,38 points from the last semester math class, raise your hand please?"



So




What is the least number of questions you can ask to guarantee to know every single student grades in the class?











share|improve this question











$endgroup$




You are a new Math 102 professor in a university class of 88 students and would like to learn their grades from previous semester to know your students better and put a note of it to get an idea how good they are because the electronic system is broken right now.



But you need the precise grades from the last semester class Math 101, which could be integers between 0 to 100. You want to ask the least number of questions not to bother your students somehow or to show off, who knows... But you can ask only one kind of question:



"How many people got x,y,z... points from the last semester math class, raise your hand please?"



and take notes of the students accordingly whoever lift his/her hands. x,y,z,... could be any number of points/grades. such as;



"How many people got 0,15,30,32,38 points from the last semester math class, raise your hand please?"



So




What is the least number of questions you can ask to guarantee to know every single student grades in the class?








mathematics logical-deduction optimization






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 6 hours ago









Community

1




1










asked 9 hours ago









OrayOray

17.2k4 gold badges39 silver badges173 bronze badges




17.2k4 gold badges39 silver badges173 bronze badges










  • 2




    $begingroup$
    I assume that "you can ask only one question" means "you can ask only one kind of question"?
    $endgroup$
    – Gareth McCaughan
    9 hours ago










  • $begingroup$
    @GarethMcCaughan fixed.
    $endgroup$
    – Oray
    8 hours ago












  • 2




    $begingroup$
    I assume that "you can ask only one question" means "you can ask only one kind of question"?
    $endgroup$
    – Gareth McCaughan
    9 hours ago










  • $begingroup$
    @GarethMcCaughan fixed.
    $endgroup$
    – Oray
    8 hours ago







2




2




$begingroup$
I assume that "you can ask only one question" means "you can ask only one kind of question"?
$endgroup$
– Gareth McCaughan
9 hours ago




$begingroup$
I assume that "you can ask only one question" means "you can ask only one kind of question"?
$endgroup$
– Gareth McCaughan
9 hours ago












$begingroup$
@GarethMcCaughan fixed.
$endgroup$
– Oray
8 hours ago




$begingroup$
@GarethMcCaughan fixed.
$endgroup$
– Oray
8 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

The answer is:




a minimum of 7 questions




Here's the way you get that number:




To solve this puzzle, I'm first going to look at a simpler variation where every student can have only 1 out of 4 possible grades: A, B, C, or D. We can do this the normal way with 3 questions (Which students got A, which students got B, which students got C. The ones with the D are the rest which didn't lift up their hand) but there's a better way.

Split up the class population in half with the question "Which students got A or B". So for every student, no matter the answer, we cut the possibility spectrum of grades they could have in half. Now, we could ask "Which students got A" to determine the grade of the people who lifted their hand in the first question, but we could also ask "Which students got A or D". If a student lifted their hand on both questions, we know they have an A because the first question told us they can't have a D. Each grade thus corresponds to an unique set of hand lifts and hand abstains, which we could represent as a binary number like so:

A - 11

B - 10

C - 00

D - 01

With every doubling of the possible number of grades, we need 1 more question to determine which half every student is in. So for 8 possible grades that is 3 questions, for 16 it's 4 and so on.
So, for N possible grades, we have to ask log2(n) questions (rounded up) to know the precise score every single student has. There are 101 scores (don't forget the 0), log2(101) is 6.65 or 7 questions needed.







share|improve this answer










New contributor



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





$endgroup$






















    5












    $begingroup$

    The answer is 7, It's a simple extension of the Magic Calculator trick.



    enter image description here



    Each card represents a bit, with the upper left index value being the value of that bit.
    To perform it as a trick, you ask the volunteer to pick a number between 1 and 64, and hand you the cards on which their number appears. Add up the index values on the selected cards, and you have the number.



    This can be extended up to 127 with one more card, indexed as 64, (and additional values on each card).



    You simply ask "Who's grade appears on card X" for each card, and record the results.



    This solution is essentially the same as @umnikos,






    share|improve this answer









    $endgroup$

















      Your Answer








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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      The answer is:




      a minimum of 7 questions




      Here's the way you get that number:




      To solve this puzzle, I'm first going to look at a simpler variation where every student can have only 1 out of 4 possible grades: A, B, C, or D. We can do this the normal way with 3 questions (Which students got A, which students got B, which students got C. The ones with the D are the rest which didn't lift up their hand) but there's a better way.

      Split up the class population in half with the question "Which students got A or B". So for every student, no matter the answer, we cut the possibility spectrum of grades they could have in half. Now, we could ask "Which students got A" to determine the grade of the people who lifted their hand in the first question, but we could also ask "Which students got A or D". If a student lifted their hand on both questions, we know they have an A because the first question told us they can't have a D. Each grade thus corresponds to an unique set of hand lifts and hand abstains, which we could represent as a binary number like so:

      A - 11

      B - 10

      C - 00

      D - 01

      With every doubling of the possible number of grades, we need 1 more question to determine which half every student is in. So for 8 possible grades that is 3 questions, for 16 it's 4 and so on.
      So, for N possible grades, we have to ask log2(n) questions (rounded up) to know the precise score every single student has. There are 101 scores (don't forget the 0), log2(101) is 6.65 or 7 questions needed.







      share|improve this answer










      New contributor



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





      $endgroup$



















        4












        $begingroup$

        The answer is:




        a minimum of 7 questions




        Here's the way you get that number:




        To solve this puzzle, I'm first going to look at a simpler variation where every student can have only 1 out of 4 possible grades: A, B, C, or D. We can do this the normal way with 3 questions (Which students got A, which students got B, which students got C. The ones with the D are the rest which didn't lift up their hand) but there's a better way.

        Split up the class population in half with the question "Which students got A or B". So for every student, no matter the answer, we cut the possibility spectrum of grades they could have in half. Now, we could ask "Which students got A" to determine the grade of the people who lifted their hand in the first question, but we could also ask "Which students got A or D". If a student lifted their hand on both questions, we know they have an A because the first question told us they can't have a D. Each grade thus corresponds to an unique set of hand lifts and hand abstains, which we could represent as a binary number like so:

        A - 11

        B - 10

        C - 00

        D - 01

        With every doubling of the possible number of grades, we need 1 more question to determine which half every student is in. So for 8 possible grades that is 3 questions, for 16 it's 4 and so on.
        So, for N possible grades, we have to ask log2(n) questions (rounded up) to know the precise score every single student has. There are 101 scores (don't forget the 0), log2(101) is 6.65 or 7 questions needed.







        share|improve this answer










        New contributor



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





        $endgroup$

















          4












          4








          4





          $begingroup$

          The answer is:




          a minimum of 7 questions




          Here's the way you get that number:




          To solve this puzzle, I'm first going to look at a simpler variation where every student can have only 1 out of 4 possible grades: A, B, C, or D. We can do this the normal way with 3 questions (Which students got A, which students got B, which students got C. The ones with the D are the rest which didn't lift up their hand) but there's a better way.

          Split up the class population in half with the question "Which students got A or B". So for every student, no matter the answer, we cut the possibility spectrum of grades they could have in half. Now, we could ask "Which students got A" to determine the grade of the people who lifted their hand in the first question, but we could also ask "Which students got A or D". If a student lifted their hand on both questions, we know they have an A because the first question told us they can't have a D. Each grade thus corresponds to an unique set of hand lifts and hand abstains, which we could represent as a binary number like so:

          A - 11

          B - 10

          C - 00

          D - 01

          With every doubling of the possible number of grades, we need 1 more question to determine which half every student is in. So for 8 possible grades that is 3 questions, for 16 it's 4 and so on.
          So, for N possible grades, we have to ask log2(n) questions (rounded up) to know the precise score every single student has. There are 101 scores (don't forget the 0), log2(101) is 6.65 or 7 questions needed.







          share|improve this answer










          New contributor



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





          $endgroup$



          The answer is:




          a minimum of 7 questions




          Here's the way you get that number:




          To solve this puzzle, I'm first going to look at a simpler variation where every student can have only 1 out of 4 possible grades: A, B, C, or D. We can do this the normal way with 3 questions (Which students got A, which students got B, which students got C. The ones with the D are the rest which didn't lift up their hand) but there's a better way.

          Split up the class population in half with the question "Which students got A or B". So for every student, no matter the answer, we cut the possibility spectrum of grades they could have in half. Now, we could ask "Which students got A" to determine the grade of the people who lifted their hand in the first question, but we could also ask "Which students got A or D". If a student lifted their hand on both questions, we know they have an A because the first question told us they can't have a D. Each grade thus corresponds to an unique set of hand lifts and hand abstains, which we could represent as a binary number like so:

          A - 11

          B - 10

          C - 00

          D - 01

          With every doubling of the possible number of grades, we need 1 more question to determine which half every student is in. So for 8 possible grades that is 3 questions, for 16 it's 4 and so on.
          So, for N possible grades, we have to ask log2(n) questions (rounded up) to know the precise score every single student has. There are 101 scores (don't forget the 0), log2(101) is 6.65 or 7 questions needed.








          share|improve this answer










          New contributor



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








          share|improve this answer



          share|improve this answer








          edited 7 hours ago





















          New contributor



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








          answered 7 hours ago









          umnikosumnikos

          1014 bronze badges




          1014 bronze badges




          New contributor



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




          New contributor




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




























              5












              $begingroup$

              The answer is 7, It's a simple extension of the Magic Calculator trick.



              enter image description here



              Each card represents a bit, with the upper left index value being the value of that bit.
              To perform it as a trick, you ask the volunteer to pick a number between 1 and 64, and hand you the cards on which their number appears. Add up the index values on the selected cards, and you have the number.



              This can be extended up to 127 with one more card, indexed as 64, (and additional values on each card).



              You simply ask "Who's grade appears on card X" for each card, and record the results.



              This solution is essentially the same as @umnikos,






              share|improve this answer









              $endgroup$



















                5












                $begingroup$

                The answer is 7, It's a simple extension of the Magic Calculator trick.



                enter image description here



                Each card represents a bit, with the upper left index value being the value of that bit.
                To perform it as a trick, you ask the volunteer to pick a number between 1 and 64, and hand you the cards on which their number appears. Add up the index values on the selected cards, and you have the number.



                This can be extended up to 127 with one more card, indexed as 64, (and additional values on each card).



                You simply ask "Who's grade appears on card X" for each card, and record the results.



                This solution is essentially the same as @umnikos,






                share|improve this answer









                $endgroup$

















                  5












                  5








                  5





                  $begingroup$

                  The answer is 7, It's a simple extension of the Magic Calculator trick.



                  enter image description here



                  Each card represents a bit, with the upper left index value being the value of that bit.
                  To perform it as a trick, you ask the volunteer to pick a number between 1 and 64, and hand you the cards on which their number appears. Add up the index values on the selected cards, and you have the number.



                  This can be extended up to 127 with one more card, indexed as 64, (and additional values on each card).



                  You simply ask "Who's grade appears on card X" for each card, and record the results.



                  This solution is essentially the same as @umnikos,






                  share|improve this answer









                  $endgroup$



                  The answer is 7, It's a simple extension of the Magic Calculator trick.



                  enter image description here



                  Each card represents a bit, with the upper left index value being the value of that bit.
                  To perform it as a trick, you ask the volunteer to pick a number between 1 and 64, and hand you the cards on which their number appears. Add up the index values on the selected cards, and you have the number.



                  This can be extended up to 127 with one more card, indexed as 64, (and additional values on each card).



                  You simply ask "Who's grade appears on card X" for each card, and record the results.



                  This solution is essentially the same as @umnikos,







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 7 hours ago









                  Chris CudmoreChris Cudmore

                  5,2171 gold badge14 silver badges39 bronze badges




                  5,2171 gold badge14 silver badges39 bronze badges






























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