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IMAC Juniori II 15 mai 2010 Subiectul III

 
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Andi Brojbeanu
Newton


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

PostPosted: Tue Jun 15, 2010 8:30 pm    Post subject: IMAC Juniori II 15 mai 2010 Subiectul III Reply with quote

Se considera 2011 numere naturale nenule a_1, a_2, ..., a_{2010}, a_{2011} si fie n=(a_1+a_2)\cdot(a_2+a_3)\cdot...\cdot(a_{2010}+a_{2011})\cdot(a_{2011}+a_1).
Sa se arate ca 2010^n-1 se divide cu 2009\cdot 2011.
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Brojbeanu Andi Gabriel, clasa XII-a
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Radu Lecoiu



Joined: 07 Sep 2017
Posts: 3

PostPosted: Thu Sep 07, 2017 3:08 pm    Post subject: Reply with quote

Vom arata mai intai ca (a_1+a_2)(a_2+a_3)...(a_{2011}+a_1) este par.
Presupunem prin reducere la absurd ca ar fi impar. Atunci toate parantezele ar fi impare.
\Rightarrow a_1 + a_2 = impar ;  a_2 + a_3 = impar ; ... a_{2011} + a_1 = impar . Deci a_i si a_{i+2} au aceeasi paritate .
Deci a_1 si a_{2011} au aceeasi paritate \Rightarrow ultimul termen al produsului va fi par . Contradictie . Presupunerea este falsa . Deci n=2k .
2010^{2k} - 1 = (2009+1)^{2k} - 1 = M_{2009} + 1^{2k} - 1 = M_{2009}
Analog 2010^{2k} - 1 = (2011-1)^{2k} - 1 = M_{2011} + 1 - 1 = M_{2011} \Rightarrow concluzia
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Radu Lecoiu
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