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Concursul "Teodor Topan" - problema 2
Posted: Mon Dec 03, 2007 2:34 pm
by maky
2. Aratati ca \( N=2002^n+2003^n+2007^n+2008^n \) se divide cu \( 5 \), oricare ar fi numarul natural impar \( n \).
Gornoava Valeriu, Zalau
Posted: Tue Mar 04, 2008 11:28 am
by mihai++
\( N=4005*(\dots)+4015*(\dots)\rightarrow N \vdots 5 \)
Posted: Tue Mar 04, 2008 8:26 pm
by Marius Dragoi
Fie \( n=2k+1 \)
\( 2002\equiv -3 (mod 5) \) \( \Rightarrow \) \( 2002^{2k+1}\equiv (-3)^{2k+1} (mod 5) \) \( \Rightarrow \) \( 2002^{2k+1}\equiv -3^{2k+1}(mod 5) \)
Analog otinem: \( 2003^{2k+1}\equiv -2^{2k+1} (mod 5) , 2007^{2k+1}\equiv 2^{2k+1} (mod 5) \) si \( 2008^{2k+1}\equiv 3^{2k+1} (mod 5) \)
Sumand obtinem: \( N\equiv 0 (mod 5) \)
Re: Concursul "Teodor Topan" - problema 2
Posted: Thu Mar 06, 2008 12:45 pm
by Virgil Nicula
2. Fie cifrele (in baza \( 10 \) ) \( a \) , \( b \) , \( c \) , \( d \) pentru care \( a+d=b+c=10 \) . Aratati ca numarul
\( N=\overline {200a}^n+\overline {200b}^n+\overline {200c}^n+\overline {200d}^n \) se divide cu \( 10 \), pentru orice numar natural impar \( n \) .
Gornoava Valeriu, Zalau
Posted: Thu Mar 06, 2008 8:50 pm
by mihai++
\( N=4010*(\dots)\rightarrow N\vdots10. \)