12 thoughts on “Pigeonhole Principle – 4

  1. This is exactly what I was searching for. I have come across similar questions and was not sure how to solve them. Thanks a lot and keep up the good work 🙂

  2. When cubes there are only 5 possible remainders… How did you find this out ? Is there a method to find this because this could be very helpful during exams. Or is it just trial and error ?

  3. Hi J! I found your post really helpful.. but I am not getting the solution of q1. If it was asked to have the minimum number of boxes with the same number of oranges, then the answer would had been 4 , am I right? Your post probably has an error in the final answer part.. 🙂

    • No, the minimum value would have been 0. There could be 0 boxes having a particular number! But this is a different question – it is asking for a minimax, that is, the minimum value of the maximum possible number. Read the earlier 3 posts again.

      Let me phrase it differently. Suppose I say, you want to find the mode of the distribution, what could be the least possible value for the mode? You will find that it could be 5 but not less.

      regards
      J

      • Sir but the minimum possible value for the mode can be 4 right? As 5×31=155. which implies that some weights must have no. of boxes less than 5. Please correct me if I’m wrong.

      • Remember that we are asked the minimum possible value of the maximum. As you said, Some may be less than 5, but do note that all cannot be less than 5. So the maximum is at least 5.

        regards
        J

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