In a mathematical competition, in which 6 problems were
posed to the participants, every two of these problems were solved by more
than 2/5 of the contestants.
Moreover, no contestant solved all the 6 problems.
Show that there are at least 2 contestants who solved exactly 5 problems
each.
source: IMO 2005
I have just made a table, based on the number of contestants (n), of how many pairs need to be solved, how many are solved if each student answers only 4 each, and the shortfall.
It looks like this:
n ***** pairs solved ***
needed if 4 each short
-- -------- ----------- ------
1 15 6 9
2 15 12 3
3 30 18 12
4 30 24 6
5 45 30 15
6 45 36 9
7 45 42 3
8 60 48 12
9 60 54 6
10 75 60 15
11 75 66 9
etc.
The 2nd column is 15*(the smallest integer greater than 2n/5)
The third column is just 6n
And the shortfall is the difference. Note that it repeats in a cycle of 5.
Since each contestant that solves five adds an additional 4 pairs of solutions (solving 10 pairs instead of 6), it follows that if n = 5k+1, at least 3 more contestants must solve 5 problems in order to get 9 additional pairs solved.
Similarly,
5k+3 requires at least 3 more contestants who solved exactly 5
5k+4 requires at least 2 more
5k+5 requires at least 4 more
So, I am all set except for 5k+2 contestants. At least one contestant needs to solve 5 in order to cover the shortfall of 3, but I do not yet see why 2 are required.
Still not there (4/5 solution?)
