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Math. competition, (Posted on 2012-03-21) Difficulty: 4 of 5
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

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Still not there (4/5 solution?) | Comment 8 of 9 |
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.  
 

  Posted by Steve Herman on 2012-03-23 08:49:55
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