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JBTST I 2010, Problema 1

 
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Andi Brojbeanu
Newton


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

PostPosted: Mon Apr 12, 2010 10:46 pm    Post subject: JBTST I 2010, Problema 1 Reply with quote

Sa se determine numerele prime p, q, r cu proprietatea ca
\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\ge 1.
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Last edited by Andi Brojbeanu on Mon Apr 26, 2010 3:57 pm; edited 1 time in total
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Alin
Euclid


Joined: 27 Mar 2010
Posts: 16

PostPosted: Mon Apr 19, 2010 8:16 pm    Post subject: Reply with quote

Pentru orice p numar prim, p\ge 5 este adevarata inegalitatea \frac{1}{p}\le \frac{1}{5}. O aplicam de 3 ori pentru p,q,r si obtinem ca \frac{1}{p}+\frac{1}{q}+\frac{1}{r}\le \frac{1}{5}+\frac{1}{5}+\frac{1}{5}<1. Deci numerele p,q,r pot fi doar 2 sau 3. Avem multimea solutiilor de triplete (p,q,r)=\left{ (2,2,2), (2,2,3), (2,3,2), (3,2,2), (2,3,3), (3,2,3), (3,3,2), (3,3,3) \right} .
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enescu
Thales


Joined: 20 May 2008
Posts: 171

PostPosted: Mon Apr 19, 2010 10:06 pm    Post subject: Reply with quote

Dar \frac12+\frac13+\frac15>1, nu?
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Laurian Filip
Site Admin


Joined: 25 Nov 2007
Posts: 473
Location: Bucuresti

PostPosted: Mon Apr 19, 2010 10:08 pm    Post subject: Reply with quote

Alin wrote:
Pentru orice p numar prim, p\ge 5 este adevarata inegalitatea \frac{1}{p}\le \frac{1}{5}. O aplicam de 3 ori pentru p,q,r si obtinem ca \frac{1}{p}+\frac{1}{q}+\frac{1}{r}\le \frac{1}{5}+\frac{1}{5}+\frac{1}{5}<1..


Ce ai aratat tu pana aici este ca cele 3 numere prime nu pot fi simultan mai mari sau egale cu 5. Acest lucru nu implica faptul ca toate cele 3 numere sunt mai mici decat 5!
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Beniamin Bogosel
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Joined: 07 Mar 2008
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PostPosted: Tue Apr 20, 2010 2:03 pm    Post subject: Reply with quote

Se poate aborda problema similar cu rezolvarea ecuatiei \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1.

Se alege o ordonare p\leq q\leq r, si se ia cel mai mic dintre p,q,r, care va trebui sa fie mai mic sau egal cu 3. Avem doua cazuri: pentru p=2,\ p=3. Mai departe, iar se alege cel mai mic dintre cele ramase, si etc.
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