# Please Help Me!

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• Feb 22nd 2006, 06:56 PM
mong sody
Please Help Me!
:mad: I can't do this exercice! Please help me!
In a mathematical competition 6 problems were posed to the contestants.Each pair of problems was solved by more than 2/5 of the constants.Nobody solved all 6 problem.Show that there were at least 2 contestants who each solved exactly 5 problems.] Thanks for your helping!!! :)
• Feb 23rd 2006, 12:36 PM
ThePerfectHacker
Quote:

Originally Posted by mong sody
:mad: I can't do this exercice! Please help me!
In a mathematical competition 6 problems were posed to the contestants.Each pair of problems was solved by more than 2/5 of the constants.Nobody solved all 6 problem.Show that there were at least 2 contestants who each solved exactly 5 problems.] Thanks for your helping!!! :)

I am slightly confused by your question. What do you mean "a pair of problems"?

Thus, far I did is this,
Let $n$ be the number of contestants. Let $x_k$ be the number of contestants which solved exactly $k$ questions. Thus,
$x_0+x_1+x_2+x_3+x_4+x_5+x_6=n$
But nobody solved all six thus, $x_6=0$. Thus we have that,
$x_0+x_1+x_2+x_3+x_4+x_5=n$.
Now, I am trying to determine which one of x's has to be greater then $\frac{2}{5}n$, but I do not understand what you mean?