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Divizibilitate cu 3

 
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gabriela_brb
Euclid


Joined: 07 Feb 2013
Posts: 12

PostPosted: Fri Oct 24, 2014 11:13 am    Post subject: Divizibilitate cu 3 Reply with quote

Buna ziua,

Va rog mult daca stiti cum se face acest exercitiu:
Aratati ca 7^a + 5^(2a+1) este divizibil cu 3.
Cu ultima cifra nu merge, cu suma cifrelor nici, pt ca avem acei exponenti.

Trebuie urgent.

Va multumesc mult de tot!
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enescu
Thales


Joined: 20 May 2008
Posts: 171

PostPosted: Fri Oct 24, 2014 9:40 pm    Post subject: Reply with quote

Daca impartim numarul 7^a la 3, ce rest obtinem?
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gabriela_brb
Euclid


Joined: 07 Feb 2013
Posts: 12

PostPosted: Sat Oct 25, 2014 7:38 pm    Post subject: Reply with quote

Va multumesc de raspuns.

Am gasit intre timp o rezolvare.

O voi posta aici pentru cine va mai da peste acest exercitiu si va vrea sa stie rezolvarea lui.

Ştim că (a+b)^n=Ma+b^n, unde ^ este ridicarea la putere, Ma este multiplu de a.

7^a=(6+1)^a=M6+1^a=M6+1

5^(2a+1)=(6-1)^(2a+1)=M6+(-1)^(2a+1)=M6-1

Dacă adunăm cele 2 relaţii de mai sus membru cu membru, avem că:

7^a + 5^(2a+1)=M6+1+M6-1=M6, deci suma din enunţ este multiplu de 6, adică se divide cu 3.
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