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JBTST II 2010, Problema 3

 
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Mon Apr 26, 2010 4:52 pm    Post subject: JBTST II 2010, Problema 3 Reply with quote

Sa se determine numerele intregi n, n\ge 2, cu proprietatea ca numerele 1!, 2!, 3!, ..., (n-1)! dau resturi diferite la impartirea cu n.
Nota: k!=1\cdot 2\cdot ...\cdot k, oricare ar fi k\ge1 natural.
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Brojbeanu Andi Gabriel, clasa XII-a
Colegiul National "Constantin Carabella" Targoviste
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Radu Lecoiu



Joined: 07 Sep 2017
Posts: 3

PostPosted: Fri Sep 08, 2017 1:49 pm    Post subject: Reply with quote

Sa observam mai intai ca n=2 este solutie .
Presupunem prin reducere la absurd ca n este compus mai mare sau egal cu 4. Atunci n=p\cdotq , q>p.
Evident q! se va divide cu n. Analog (q+1)!. \Rightarrowq+1 nu face parte din sirul 1,2,...n-1. Deci n-1=q. n-1|n \Rightarrown=2. Contradictie. Presupunerea facuta este falsa.
Deci n este prim.
Evident 1! va da restul 1 la impartirea cu n
Cum n este prim , din teorema lui Wilson avem ca (p-1)!\equiv p-1 (mod p)
Adica (p-2)!\equiv 1 (mod p). Pentru a nu contrazice ipoteza obtinem p-2=1
p=3. Deci , singurele solutii sunt 2 si 3.
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Radu Lecoiu
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Colegiul National "Ienachita Vacarescu"
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