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A trivial but efficient numerical proof without assuming that friends of my friends are my friends:

Each pair know a third fellow. There are \(\binom 5 2 = 10\)  pairs \(\equiv 20\)  arrows starting from each to another element ( no self arrow allowed ).

The mean of received arrows is \(2 \frac{\binom 5 2}5 = 2 \frac{10}5 = 4\)

Receiving 4 arrows means the receiver knows the other \(4 \equiv\)  everybody.

It is a mean. If you decrease the arrows received by someone , you must increase the arrows received by another one.

Hence, there is always someone receiving at least 4 arrows. Then there is always someone which is the friend of "everybody".
 

question 1 Juillet 15 par ladmin (9 points)  
editée 7 Octobre 16 par ladmin
formule dans les commentaires ? \( x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
ok
mais il faut la préparer et l'insérer ici entre 'antislash( '  et 'antislash) '  en remplaçant  l'anti-slash par un toucher de clavier standard sur  Alt Gr 8
Alors qu'elle est en wysiwyg dans les questions réponses
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