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
Concursul "Teodor Topan" - problema 2
Moderators: Bogdan Posa, Laurian Filip
-
Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
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: \( N\equiv 0 (mod 5) \)
\( 2002\equiv -3 (mod 5) \) \( \Rightarrow \) \( 2002^{2k+1}\equiv (-3)^{2k+1} (mod 5) \) \( \Rightarrow \) \( 2002^{2k+1}\equiv -3^{2k+1}(mod 5) \)
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: \( N\equiv 0 (mod 5) \)
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Re: Concursul "Teodor Topan" - problema 2
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
\( N=\overline {200a}^n+\overline {200b}^n+\overline {200c}^n+\overline {200d}^n \) se divide cu \( 10 \), pentru orice numar natural impar \( n \) .
Gornoava Valeriu, Zalau