Code: Select all
A=(2004a+2005c)/(2005b+2004c);
B=(2004b+2005c)/(2005c+2004a);
C=(2004c+2005b)/(2005c+2004b);Fractii
Moderators: Bogdan Posa, Laurian Filip
O sa restructurez datele problemei
\( A=\frac{2004a+2005c}{2005b+2004c} \)
\( B=\frac{2004b+2005c}{2005c+2004a} \)
\( C=\frac{2004c+2005b}{2005c+2004b} \)
Cred ca iese destul de usor cu divizibiliate.
Cum \( A \in Z => (2005b+2004c) | (2004a+2005c) \)
\( B \in Z => (2005c+2004a) | (2004b+2005c) \)
\( C \in Z => (2005c+2004b) | (2004b+2005c) \)
Folosim proprietatea divizibilitatii \( a|a \)
\( => (2005b+2004c) | (2005b+2004c) \)
\( => (2005c+2004a) | (2005c+2004a) \)
\( => (2005c+2004b) | (2005c+2004b) \)
Scadem relatiile s.a.m.d. si ar trebui sa rezulte niste solutii simetrice de tipul
\( a=1 \)
\( b=1 \)
\( c=1 \)
\( A=\frac{2004a+2005c}{2005b+2004c} \)
\( B=\frac{2004b+2005c}{2005c+2004a} \)
\( C=\frac{2004c+2005b}{2005c+2004b} \)
Cred ca iese destul de usor cu divizibiliate.
Cum \( A \in Z => (2005b+2004c) | (2004a+2005c) \)
\( B \in Z => (2005c+2004a) | (2004b+2005c) \)
\( C \in Z => (2005c+2004b) | (2004b+2005c) \)
Folosim proprietatea divizibilitatii \( a|a \)
\( => (2005b+2004c) | (2005b+2004c) \)
\( => (2005c+2004a) | (2005c+2004a) \)
\( => (2005c+2004b) | (2005c+2004b) \)
Scadem relatiile s.a.m.d. si ar trebui sa rezulte niste solutii simetrice de tipul
\( a=1 \)
\( b=1 \)
\( c=1 \)
Last edited by Al3xx on Tue Feb 17, 2009 12:23 pm, edited 1 time in total.
-
Virgil Nicula
- Euler
- Posts: 622
- Joined: Fri Sep 28, 2007 11:23 pm
Re: Fractii
MariusG, invata si tu LaTeX - ul. Era chiar mai simplu de scris. Nu mai vorbesc de "frumos".MariusG wrote:Se dau fractiile:Determinati a,b,c in Z stiind ca A,B,C sunt simultan numere intregiCode: Select all
A=(2004a+2005c)/(2005b+2004c); B=(2004b+2005c)/(2005c+2004a); C=(2004c+2005b)/(2005c+2004b);
MariusG wrote:Se dau fractiile \( A=\frac {2004a+2005c}{2005b+2004c}\ ,\ B=\frac {2004b+2005c}{2005c+2004a}\ ,\ C=\frac {2004c+2005b}{2005c+2004b}\ . \)
Determinati numerele intregi \( a\ ,\ b\ ,\ c \) stiind ca \( A ,\ B\ ,\ C \) sunt simultan numere intregi.
Mersi dar am incercat si eu dar nu a mers.Al3xx wrote:O sa restructurez datele problemei
\( A=\frac{2004a+2005c}{2005b+2004c} \)
\( B=\frac{2004b+2005c}{2005c+2004a} \)
\( C=\frac{2004c+2005b}{2005c+2004b} \)
Cred ca iese destul de usor cu divizibiliate.
Cum \( A \in Z => (2005b+2004c) | (2004a+2005c) \)
\( B \in Z => (2005c+2004a) | (2004b+2005c) \)
\( C \in Z => (2005c+2004b) | (2004b+2005c) \)
Folosim proprietatea divizibilitatii \( a|a \)
\( => (2005b+2004c) | (2005b+2004c) \)
\( => (2005c+2004a) | (2005c+2004a) \)
\( => (2005c+2004b) | (2005c+2004b) \)
Scadem relatiile s.a.m.d. si ar trebui sa rezulte niste solutii simetrice de tipul
\( a=1 \)
\( b=1 \)
\( c=1 \)
E undeva o lista cu comenzi LaTeX?
Marius Gavrilescu