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lgbasallote Group Title

show that at least 3 of any 25 days chosen must fall in the same month of the year

  • 2 years ago
  • 2 years ago

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  1. CliffSedge Group Title
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    Including leap years?

    • 2 years ago
  2. bahrom7893 Group Title
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    Simple pigeonhole principle.

    • 2 years ago
  3. bahrom7893 Group Title
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    Only 12 months, 25 days

    • 2 years ago
  4. CliffSedge Group Title
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    What if the 25 days span two years?

    • 2 years ago
  5. bahrom7893 Group Title
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    He means 25 days of the same year. Otherwise this would not be true.

    • 2 years ago
  6. bahrom7893 Group Title
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    Suppose you try to distribute them as evenly as you can with the minimal amount of days per month, that would be: 24/12 = 2 days per month, but you have 1 more day left over.

    • 2 years ago
  7. bahrom7893 Group Title
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    So whatever month you add it to, that month will end up with exactly 3 days.

    • 2 years ago
  8. bahrom7893 Group Title
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    if you decide to give a certain month only 1 day, then you'll have to have 4 days in a different month.

    • 2 years ago
  9. bahrom7893 Group Title
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    It's basically saying that if I have two boxes and 3 balls, and I have to put those balls into the boxes, one box is going to have more than 2 balls

    • 2 years ago
  10. bahrom7893 Group Title
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    I meant more than 1 ball

    • 2 years ago
  11. lgbasallote Group Title
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    it's proving...not a question

    • 2 years ago
  12. lgbasallote Group Title
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    anyway...

    • 2 years ago
  13. lgbasallote Group Title
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    what were you saying @bahrom7893 ?

    • 2 years ago
  14. bahrom7893 Group Title
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    Well i just proved it. Consider the worst case scenario, where you would want to distribute the days into the months, so that each month has the minimum amount of days. Put one day into each month, that would give you 12 months, with one day in each, and you have 13 days remaining to be distributed. Repeat - add a day to each month, that would leave you with 12 months, with two days in each, and you still have 1 day left to be distributed.

    • 2 years ago
  15. bahrom7893 Group Title
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    Since the number of days in each month is already 2, whichever month you decide to put the last day in, is going to have 3 days in it.

    • 2 years ago
  16. bahrom7893 Group Title
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    And that is the proof.

    • 2 years ago
  17. lgbasallote Group Title
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    hmm that does make sense...

    • 2 years ago
  18. bahrom7893 Group Title
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    if you decide to give 0 days to one month, you're already screwed because you have more days left. So in these types of problems you'd want to distribute a minimum amount of whatever you're distributing, evenly among the boxes.

    • 2 years ago
  19. bahrom7893 Group Title
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    and then see if you have any leftovers.

    • 2 years ago
  20. lgbasallote Group Title
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    well ot's at;east...so i suppose even if you assign 0 days to one month...it will still make one month have 3 days

    • 2 years ago
  21. bahrom7893 Group Title
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    it will have one month have 4 days and one month to haave 3 days i believe

    • 2 years ago
  22. lgbasallote Group Title
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    it still fits the condition right?

    • 2 years ago
  23. bahrom7893 Group Title
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    yea, if you decrease the number of months or increase the number of days, it will fit the condition. If you decrease the number of days, or increase the number of months (which is silly) then it won't.

    • 2 years ago
  24. lgbasallote Group Title
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    so no matter what happens...you're right?

    • 2 years ago
  25. bahrom7893 Group Title
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    yea, for example, put 0 days in one month. That would be pretty much the same as getting rid of that month. So now you have to distribute 25 days into 11 months. Do what I said. Step 1: 11 months, one day in each, 14 days remaining Step 2: 11 months, two days in each, 3 days remaining So now you have 3 additional months, and whichever month you add them to, is going to have at least 3 days.

    • 2 years ago
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