Show that at least 3 of any 25 days chosen must fall in the same month of the year.
Answer (3)
==> In order to figure this out, you have to know that there are 12 months in the year.
Take a calendar. You have to pick 25 days in it.
First 12 days you pick: You want to avoid bunching them up, so you pick each one from a different month.
Next 12 days you pick: You want to avoid bunching them up, so you pick each of these from a different month too.
So far, you've picked (12 + 12) = 24 numbers, and there are two from each month.
Last day you pick: You still have to pick one more number to make 25. What month will you pick it from ? WHATEVER month you pick it from, you'll have three days from that month.
To show that at least 3 of any 25 days chosen must fall in the same month, we can use the Pigeonhole Principle. The principle essentially states that if you have n pigeonholes (or buckets) to fill and m items to distribute, where m > n, at least one pigeonhole must contain more than one item. In the case of days and months, there are 12 months in a year. If we choose 25 days, we can think of them as items to be distributed into the 12 pigeonholes (the months).
Using the most simplified scenario, distributing the days evenly, the most any month would get is 2 days (24/12=2). However, since we have 25 days to allocate, one additional day will be left over, and it must be placed into one of the already filled months, resulting in at least one month having 3 days.
This conclusion is based on the fact that most months have 30 or 31 days, with the exception of February, which has 28 days in a regular year and 29 in a leap year.
By applying the Pigeonhole Principle, we can determine that with 25 selected days and only 12 months available, at least three of those days must fall in the same month. If every month had only two days, it totals to a maximum of 24 days, making the 25th day land in an already chosen month. Therefore, it's guaranteed that at least three days will be from the same month.
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