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 

divizor

 
Post new topic   Reply to topic    mateforum.ro Forum Index -> Clasa a V-a
View previous topic :: View next topic  
Author Message
Andreea Udrea
Euclid


Joined: 23 Oct 2008
Posts: 12

PostPosted: Thu Dec 04, 2008 7:02 pm    Post subject: divizor Reply with quote

Aratati ca numarul 76^63+66^63 se divide la 71.
Back to top
View user's profile Send private message
Marius Dragoi
Thales


Joined: 31 Jan 2008
Posts: 126
Location: Bucharest

PostPosted: Fri Dec 05, 2008 4:25 pm    Post subject: Reply with quote

76 \equiv 5 (mod 71) \Rightarrow 76^{63} \equiv 5^{63} (mod 71) (1)
66 \equiv -5 (mod 71) \Rightarrow 66^{63} \equiv -{5^{63}} (mod 71) (2)
Din (1) si (2) \Rightarrow  76^{63}+66^{63} \equiv 0 (mod 71) QED Wink
_________________
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
Back to top
View user's profile Send private message Yahoo Messenger
Marcelina Popa
Bernoulli


Joined: 05 Mar 2008
Posts: 208
Location: Tulcea

PostPosted: Fri Dec 05, 2008 11:07 pm    Post subject: Reply with quote

Alta rezolvare, tot la nivelul clasei IX-a: demonstram prin inductie ca
76^{2n+1}+66^{2n+1}\ \vdots \ 71,
pentru orice numar natural n.

Mai merge si cu formula binomului lui Newton (clasa a X-a).

Nu cred ca exista vreo rezolvare la nivelul clasei a V-a. Poate doar vreo simulare de inductie.
Back to top
View user's profile Send private message Visit poster's website
Petrescubianka



Joined: 21 Feb 2011
Posts: 3

PostPosted: Tue Feb 22, 2011 10:59 am    Post subject: Reply with quote

nici yo nu ma descurcam cu astea Very Happy
Back to top
View user's profile Send private message
opincariumihai
Thales


Joined: 09 May 2009
Posts: 142
Location: BRAD

PostPosted: Tue Feb 22, 2011 9:32 pm    Post subject: Reply with quote

La nivelul clase a V-a se considera cunoscut faptul ca  a^n+b^n este multiplu de a+b in cazul n impar . Cum 66+76=2*71se obtine imediat concluzia.
Back to top
View user's profile Send private message Send e-mail
seby97
Euclid


Joined: 04 Aug 2011
Posts: 31

PostPosted: Tue Aug 16, 2011 12:39 pm    Post subject: Reply with quote

76=71+5 iar 66=71-5
acuma ecuatia va deveni (71+5)^63 +(71-5)^63
Vom utiliza formula:(a+b)^n=Multiplu de a + b^n si (a-b)^2=Multiplu de a +(-b)^2
Avem M71 +5^63 + M71 - 5^63=2M71=M71
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