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Ecuatie aparent simpla
Posted: Wed Mar 04, 2009 4:16 pm
by BurnerD1
Gasiti valorile numerelor \( x,y \in \mathbb{N} \) care satisfac ecuatia
\( xy + 6(x+y)=1987 \)
Posted: Wed Mar 04, 2009 5:36 pm
by DrAGos Calinescu
Se obtine \( x=\frac{1987-6y}{y+6}\Longrightarrow y+6/1987-6y\Longrightarrow y+6/2023 \)
\( 2023=7\cdot 17^2 \)
Deci \( y+6\in {7,17,119,289,2023} \)
La fel si \( x+6 \) e in aceeasi multime si verifici ce valori satisfac relatia.
nu inteleg..
Posted: Wed Mar 04, 2009 5:45 pm
by BurnerD1
Nu prea inteleg... cum din \( y+6 | 1987 - 6y \) rezulta ca \( y+6 | 2023 \) ??
Posted: Wed Mar 04, 2009 6:12 pm
by DrAGos Calinescu
\( y+6/1987-6y \)
Dar \( y+6/6y+36 \)
Stim ca daca \( a/b \) si \( a/c \Longrightarrow a/b+c \)
Deci din cele doua relatii \( y+6/2023 \)
Posted: Wed Mar 04, 2009 6:24 pm
by BurnerD1
deci daca \( y+6 | 1987, y+6 | 6y+36 \) atunci \( y+6 \) nu divide \( 2023 + 6y? \)
Posted: Wed Mar 04, 2009 6:31 pm
by DrAGos Calinescu
Scuze am modificat era\( 1987-6y \)
Posted: Wed Mar 04, 2009 7:16 pm
by BurnerD1
Mai redacteaza te rog inca o data rezolvarea, fara greseli
Posted: Wed Mar 04, 2009 7:17 pm
by DrAGos Calinescu
E corecta, al doilea post avea o greseala.
Posted: Wed Mar 04, 2009 7:20 pm
by BurnerD1
eu iti spun ca nu e corecta.. tu ai scris ca daca \( y+6|1987 \Longrightarrow y+6|2023 \). Trebuia \( y+6 | 2023 + 6y \). Am dreptate sau nu?
Posted: Wed Mar 04, 2009 7:22 pm
by DrAGos Calinescu
Nu...\( y+6/2023 \)
Posted: Wed Mar 04, 2009 7:49 pm
by Al3xx
Sa nu o mai lungim atata ...
\( y+6|1987-6y(1)
y+6|y+6 \rightarrow y+6|6(y+6) \rightarrow y+6|6y+36(2) \)
Adunand (1) si (2) \( \rightarrow y+6|1987-6y+6y+36<=> y+6|2023 \)
deci e corect ..
Posted: Wed Mar 04, 2009 9:15 pm
by BurnerD1
Scuze, acum am observat, mc mult!