lgbasallote
  • lgbasallote
show that at least 3 of any 25 days chosen must fall in the same month of the year
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Including leap years?
bahrom7893
  • bahrom7893
Simple pigeonhole principle.
bahrom7893
  • bahrom7893
Only 12 months, 25 days

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anonymous
  • anonymous
What if the 25 days span two years?
bahrom7893
  • bahrom7893
He means 25 days of the same year. Otherwise this would not be true.
bahrom7893
  • bahrom7893
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.
bahrom7893
  • bahrom7893
So whatever month you add it to, that month will end up with exactly 3 days.
bahrom7893
  • bahrom7893
if you decide to give a certain month only 1 day, then you'll have to have 4 days in a different month.
bahrom7893
  • bahrom7893
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
bahrom7893
  • bahrom7893
I meant more than 1 ball
lgbasallote
  • lgbasallote
it's proving...not a question
lgbasallote
  • lgbasallote
anyway...
lgbasallote
  • lgbasallote
what were you saying @bahrom7893 ?
bahrom7893
  • bahrom7893
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.
bahrom7893
  • bahrom7893
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.
bahrom7893
  • bahrom7893
And that is the proof.
lgbasallote
  • lgbasallote
hmm that does make sense...
bahrom7893
  • bahrom7893
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.
bahrom7893
  • bahrom7893
and then see if you have any leftovers.
lgbasallote
  • lgbasallote
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
bahrom7893
  • bahrom7893
it will have one month have 4 days and one month to haave 3 days i believe
lgbasallote
  • lgbasallote
it still fits the condition right?
bahrom7893
  • bahrom7893
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.
lgbasallote
  • lgbasallote
so no matter what happens...you're right?
bahrom7893
  • bahrom7893
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.

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