The Pigeon Hole Principle

Suppose a musical group has 11 weeks to prepare for opening night, and
they intend to have at least one rehearsal each day.  However, they
decide not to schedule more than 12 rehearsals in any 7-day period,
to keep from getting burned out.  Prove that there exists a sequence
of successive days during which the band has exactly 21 rehearsals.

This problem can be solved by a double application of the pigeonhole 
principle.  First, since the band has no more than 12 rehearsals
per week, they can't have more than 132 rehearsals in 11 weeks.  (To
see this, let x_k denote the number of games played in each of the 11 
consecutive weeks, and note that they cannot sum to greater than 132 
if they are each no greater than 12.)

Now let x_n denote the total number of rehearsals that have been held
after n days. Since the band rehearses at least once per day, we have

      1  <=  x_1  <  x_2  <  x_3  <  ...  <  x_77  <=  132

Also, we can add 21 to each of these numbers to give the sequence

    (x_1 + 21)  <  (x_2 + 21)  < ...  <  (x_77 + 21)  <=  153

There are 77 numbers x_n and 77 more numbers x_n + 21 for a total of
154 numbers, all in the range from 1 to 153.  Thus, at least two of
these 154 numbers must be equal.  But the x_n are all distinct, as
are the (x_n + 21), so any "overlap" must be between a number of the
form x_n and one of the form (x_m + 21), which implies 

                   x_n  =  x_m + 21

This proves that there are indices m,n such that exactly 21 rehearsals
were held during the sequence of consecutive days from day n+1 to
day m.