mateforum.ro Forum Index mateforum.ro

 
 FAQFAQ   SearchSearch   MemberlistMemberlist   UsergroupsUsergroups   RegisterRegister 
 ProfileProfile   Log in to check your private messagesLog in to check your private messages   Log inLog in 

Divizibilitate cu 13

 
Post new topic   Reply to topic    mateforum.ro Forum Index -> Clasa a V-a
View previous topic :: View next topic  
Author Message
Andi Brojbeanu
Newton


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

PostPosted: Sun Apr 19, 2009 5:19 pm    Post subject: Divizibilitate cu 13 Reply with quote

Sa se afle numarul natural n cel mai apropiat de 1997 pentru care 3+3^2+....+3^n se divide cu 13.

Probleme date la olimpiade, RMT 1/1998
Back to top
View user's profile Send private message Send e-mail Yahoo Messenger
salazar
Thales


Joined: 06 Apr 2009
Posts: 100
Location: Alba Iulia

PostPosted: Sun Apr 19, 2009 8:11 pm    Post subject: Reply with quote

S=3+3^2+...+3^n
3S=3^2+3^3+...+3^n+1
2S=3^n+1-3
S=\frac{3(3^n-1)}{2}
3(3^n-1)=26k
pt.n=1998,3^1^9^9^8-1=26k
(3^3)^6^6^6-1=26k
27^6^6^6-1=26k
(26+1)^6^6^6-1=26k
M_2_6+1-1=26k
M_2_6=26kADEVARAT
observem ca pt.n divizibil cu 27, avem M_2_6+1-1+26k\Longrightarrown trebuie sa fie divizibil cu 27,deci puterea lui divizibila cu 3.Cel mai apropiat numar de 1997 care respecta cerinta este 1998.
Back to top
View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    mateforum.ro Forum Index -> Clasa a V-a All times are GMT + 2 Hours
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum



Powered by phpBB © 2001, 2005 phpBB Group