The grades of the students in a classPast, 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

What parameters are to be considered when choosing a MOSFET?

How do I safety check that there is no light in Darkroom / Darkbag?

Numerically Stable IIR filter

Create two random teams from a list of players

How does Asimov's second law deal with contradictory orders from different people?

Why didn't General Martok receive discommendation in Star Trek: Deep Space Nine?

What is the range of a Drunken Monk's Redirect attack?

Why does macOS create file mounts for each app?

A conjectural trigonometric identity

Applications of pure mathematics in operations research

Can birds evolve without trees?

Is it unprofessional to mention your cover letter and resume are best viewed in Chrome?

Betrayed by management at a new job, should I take action?

Easy way to get process information from a window

What is my clock telling me to do?

May a hotel provide accommodation for fewer people than booked?

In the Schrödinger equation, can I have a Hamiltonian without a kinetic term?

How do I respond appropriately to an overseas company that obtained a visa for me without hiring me?

What kind of horizontal stabilizer does a Boeing 737 have?

Derivative is just speed of change?

How can a class have multiple methods without breaking the single responsibility principle

Using Python in a Bash Script

If the Moon were impacted by a suitably sized meteor, how long would it take to impact the Earth?

Can I shorten this filter, that finds disk sizes over 100G?



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








      StackExchange.ready(function()
      var channelOptions =
      tags: "".split(" "),
      id: "559"
      ;
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function()
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled)
      StackExchange.using("snippets", function()
      createEditor();
      );

      else
      createEditor();

      );

      function createEditor()
      StackExchange.prepareEditor(
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      imageUploader:
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      ,
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      );



      );













      draft saved

      draft discarded


















      StackExchange.ready(
      function ()
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f86788%2fthe-grades-of-the-students-in-a-class%23new-answer', 'question_page');

      );

      Post as a guest















      Required, but never shown

























      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






























                      draft saved

                      draft discarded
















































                      Thanks for contributing an answer to Puzzling Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid


                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.

                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function ()
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f86788%2fthe-grades-of-the-students-in-a-class%23new-answer', 'question_page');

                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

                      Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

                      Tom Holland Mục lục Đầu đời và giáo dục | Sự nghiệp | Cuộc sống cá nhân | Phim tham gia | Giải thưởng và đề cử | Chú thích | Liên kết ngoài | Trình đơn chuyển hướngProfile“Person Details for Thomas Stanley Holland, "England and Wales Birth Registration Index, 1837-2008" — FamilySearch.org”"Meet Tom Holland... the 16-year-old star of The Impossible""Schoolboy actor Tom Holland finds himself in Oscar contention for role in tsunami drama"“Naomi Watts on the Prince William and Harry's reaction to her film about the late Princess Diana”lưu trữ"Holland and Pflueger Are West End's Two New 'Billy Elliots'""I'm so envious of my son, the movie star! British writer Dominic Holland's spent 20 years trying to crack Hollywood - but he's been beaten to it by a very unlikely rival"“Richard and Margaret Povey of Jersey, Channel Islands, UK: Information about Thomas Stanley Holland”"Tom Holland to play Billy Elliot""New Billy Elliot leaving the garage"Billy Elliot the Musical - Tom Holland - Billy"A Tale of four Billys: Tom Holland""The Feel Good Factor""Thames Christian College schoolboys join Myleene Klass for The Feelgood Factor""Government launches £600,000 arts bursaries pilot""BILLY's Chapman, Holland, Gardner & Jackson-Keen Visit Prime Minister""Elton John 'blown away' by Billy Elliot fifth birthday" (video with John's interview and fragments of Holland's performance)"First News interviews Arrietty's Tom Holland"“33rd Critics' Circle Film Awards winners”“National Board of Review Current Awards”Bản gốc"Ron Howard Whaling Tale 'In The Heart Of The Sea' Casts Tom Holland"“'Spider-Man' Finds Tom Holland to Star as New Web-Slinger”lưu trữ“Captain America: Civil War (2016)”“Film Review: ‘Captain America: Civil War’”lưu trữ“‘Captain America: Civil War’ review: Choose your own avenger”lưu trữ“The Lost City of Z reviews”“Sony Pictures and Marvel Studios Find Their 'Spider-Man' Star and Director”“‘Mary Magdalene’, ‘Current War’ & ‘Wind River’ Get 2017 Release Dates From Weinstein”“Lionsgate Unleashing Daisy Ridley & Tom Holland Starrer ‘Chaos Walking’ In Cannes”“PTA's 'Master' Leads Chicago Film Critics Nominations, UPDATED: Houston and Indiana Critics Nominations”“Nominaciones Goya 2013 Telecinco Cinema – ENG”“Jameson Empire Film Awards: Martin Freeman wins best actor for performance in The Hobbit”“34th Annual Young Artist Awards”Bản gốc“Teen Choice Awards 2016—Captain America: Civil War Leads Second Wave of Nominations”“BAFTA Film Award Nominations: ‘La La Land’ Leads Race”“Saturn Awards Nominations 2017: 'Rogue One,' 'Walking Dead' Lead”Tom HollandTom HollandTom HollandTom Hollandmedia.gettyimages.comWorldCat Identities300279794no20130442900000 0004 0355 42791085670554170004732cb16706349t(data)XX5557367