Show that at least 4 of any 22 dates in the calendar must fall on the same day of the week

Question

Show that at least 4 of any 22 dates in the calendar must fall on the same day of the week

I have a question regarding this proof. When we assume the opposite is true, that is, assume that no more than 3 of the 22 days fall on the same day of the week. Does this mean 22 consecutive days or 22 random dates? Or does it mean that we have to choose 3 full weeks + 1 extra day which would obviously contradict the assumption that 22 days were picked so 21 days were picked.

If any 21 random dates were picked say... 26th july-31st july (6 days), 2nd august-7th august (6 days), 9th august-14th august (6 days), 16th august-18th august (3 days) = 21 days then 4 mondays can be picked, 4 tuesdays or 4 wednesdays so I am confused by how picking 21 days results in at most 3 of days falling on the same day of the week.

So does it mean any 21 days or 21 consecutive days?

Answer 1

It means $22$ random dates. And nobody claims that picking $21$ days always results in at most $3$ of days falling on the same day of the week. But sometimes it does. Just pick $3$ random Saturdays, $3$ random Sundays, and so on. In the end, you will have $21$ says, with exactly $3$ of days falling on each day of the week.

Answer 2

I am afraid I cannot see which part of your question is the proof that you want to see explained. If you could clarify where the unclear proof starts and ends in your post, we can address your actual question.

However, if you are also happy with a reference so that you can solve the problem on your own, then you should have a look at the pigeonhole principle: https://en.wikipedia.org/wiki/Pigeonhole_principle The generalized version applies to your problem. The pigeon holes are the seven days of the week, and the 22 dates are the pigeons. If you don't believe the pigeonhole principle, try to prove it by contradiction.

Edit: After reading some of your comments, I think that the following might be the problem. The proof states: If you pick days from the calendar such that each day of the week appears $\leqslant 3$ times, then you picked only up to $21$ days.
Maybe you understood it like: If I pick $21$ days from the calendar, then each day of the week occurs $\leqslant 3$ times.

The first reading is intended in the proof, and the statement is true. The second understanding of the proof is indeed wrong, as you pointed out in several comments (with the counterexample of 21 Wednesdays).

Answer 3

The idea is to use $7(3)+1=22$ to show there are at least $3+1=4$ dates on at least 1 day of the week. Even split $21$ allows $3$ each day of the week, and in any other case by law of averages, at least $1$ day has more than $3$ . So assume the exceptional case for $21$, then adding $1$ date, forces at least $4$ on exactly one day of the week.