There are odd number of questions in an exam:

  • Every question is answered correctly by at least one of the students in the class.
  • Every student answers correctly an even number of questions.
  • The number of questions correctly answered by any two students is even.

What is the minimum number of questions with the given condition above?

  • $\begingroup$ the number of questions answered or the number assigned to the question asked?? Ie A answers questions 2 and 3- is the value you're looking for 2 or 5? $\endgroup$ – Jason V Aug 8 '17 at 2:09
  • $\begingroup$ Third point do not add any significant information as it can be written as "The number of questions correctly answered by any number of students is even." $\endgroup$ – Nikhil Bhavar Aug 8 '17 at 7:15
  • $\begingroup$ @NikhilBhavar without third info, the answer would be 3. $\endgroup$ – Oray Aug 8 '17 at 7:25
  • $\begingroup$ Second point says, every student answers even number of questions. And we know sum of odd/even number of even numbers is even. Which makes the third point look redundant. $\endgroup$ – Nikhil Bhavar Aug 8 '17 at 7:39
  • $\begingroup$ @NikhilBhavar I believe you are wrong, image there are two students with 3 questions, first student answer 1. and 2. question correctly, second student answer 2. and 3. question. So both students answered correctly even number of questions. and 3 questions would be enough. I believe you need to read the question and the answer below. $\endgroup$ – Oray Aug 8 '17 at 7:42

Minimum questions is:



Student 1 answers 1, 2, 3, and 4 correctly
Student 2 answers 3, 4, 5, and 6 correctly
Student 3 answers 1, 3, 5, and 7 correctly

✔ Number of questions is odd.
✔ Each question is answered correctly at least once.
✔ Every student answers four correctly; 4 is an even number.
✔ Any two students answer an even number of questions -
  1+2 answer 1, 2, 3, 4, 5, and 6 - six questions is even.
  1+3 answer 1, 2, 3, 4, 5, and 7 - six questions is even.
  2+3 answer 1, 3, 4, 5, 6, and 7 - six questions is even.

It can't be less, because -

The number of questions must be odd.
• It can't be 1, as then the only even number of questions students could answer would be zero, and if all answer zero, not all questions get answered.
• It can't be 3, as then there are only three combinations of an even number of correct answers available to any student (we exclude "0"): 1+2, 1+3, and 2+3. At least two of these are needed to cover all 3 questions, and there is no way to pick two without there being a pair of students whose combined answers is 3, and thus not even.
• It can't be 5; each student could have two or four answers, but any combination of answerers that includes a student with four correct answers means that student plus a student with the fifth answer would make those two students a pair whose combined answers is 5, and thus not even - so each student can answer at most 2. You then run into a parallel scenario as for 3 questions, where it is always possible to find at least one pair of two answers out of any set of pairs that covers all of 1 to 5 that overlaps on one value, resulting in a pair of students whose combined answers is 3, and thus not even.

  • $\begingroup$ Technically, this answer could be considered incorrect as the statement "The number of students is unknown." is not obviously true. But if you added an alternative solution with 6 students (student n and student n+3 could both have the same answers for example) and seven questions, you would be right. $\endgroup$ – Penguino Aug 7 '17 at 23:50
  • $\begingroup$ Oray has apparently removed that statement now. :) $\endgroup$ – Rubio Aug 8 '17 at 1:27

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