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 Math. competition, (Posted on 2012-03-21)
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)

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 Is there an error of ways? | Comment 3 of 9 |
The text of the problem is somewhat unclear and abiguous.
Does "every two of these problems were solved by more than 2/5 of the contestants" mean that more than 2/5 of the contestants solved at least two problems, or does it mean that at more than 2/5 of the contestants solved at least 3, i.e. half of the problems. If the meaning is either of the two, and as there is no restriction on any participant solving the same number of problems (except that all 6 were not solved by any one contestant), it can not be definitively shown that 2 contestants solved exactly 5 problems each, as it would be possible that all participants could have solved the same 2 or 3 problems.
Can you clarify the criteria of the problem?
 Posted by Dej Mar on 2012-03-21 14:35:26

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