(20 points) On a certain day in a certain month in 2001 at midnight, teachers and students of a certain school were watching the live broadcast from Moscow of the announcement of the host city for the 2008 Summer Olympic Games. Beijing won the bid. In their excitement, they noticed that the number of people present was the day number, the number of boys was the month number, and the numbers of teachers, students, month, and day were all prime numbers. The number of boys was the highest, and the number of teachers was the lowest. After calculation, the sum of the number of students, the month number, and the day number, minus the number of teachers, was equal to the number of the 2008 Olympic Games, which was no more than 30. Try to answer:
(1) On which day and month did Beijing win the bid for the 2008 Olympic Games?
(2) Which edition of the Olympics was the 2008 Games?
Let the number of teachers be $a$, the number of students be $b$, the number of months be $c$, the number of days be $d$, and the number of girls be $e$. From the problem, we have
$$
\left\{\begin{array}{l}
a+b=d, \\
c+p=b . \\
b+c+d-a \leqslant 30, \\
au$, so $e$ is even.
Substituting (1) into (1), we get $2 b+c \leqslant 30$.
Since $c<b$, we have $3 c<2 b+c \leqslant 30$.
Given $c<10$, we know $c=3,5,7$.
When $c=3$, from $2<e<3$, there is no solution.
When $c=5$, from $2<p<5, e=4$, and $b=9$, which does not meet the problem's conditions.
When $c=7$, from $2<e<7, e=4$ or $e=6$. If $e=6$, then $b=13, d=15$. This does not meet the problem's conditions. If $e=4$, then $b=11, d=13$.
In summary, $a=2, b=11, c=7, d=13$.
Thus, $b+c+d-a=29$.
Answer: Beijing won the right to host the 2008 Olympic Games on July 13, 2001, and the 2008 Olympics was the 29th.