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Thread: Please Help Me!

  1. #1
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    Angry Please Help Me!

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