Fie \( f,g \in \mathbb{Z}[X] \) neconstante astfel incat \( g | f \) in \( \mathbb{Z}[X] \). Demonstrati ca daca polinomul \( f-2008 \) are cel putin 81 de radacini intregi, atunci gradul lui \( g \) este mai mare decat 5.
IMC 2008
IMC 2008 ziua 2 problema 4
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Beniamin Bogosel
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IMC 2008 ziua 2 problema 4
Yesterday is history,
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog
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Beniamin Bogosel
- Co-admin
- Posts: 710
- Joined: Fri Mar 07, 2008 12:01 am
- Location: Timisoara sau Sofronea (Arad)
- Contact:
Scriem \( f=gh \) cu \( h \in \mathbb{Z}[X] \) si astfel exista \( x_1,...,x_{81} \) astfel incat \( g(x_i)|2008,\ i=1,..,81 \). Cum 2008 are 16 divizori intregi, exista sase dintre \( x_i \) astfel incat \( g \) ia aceeasi valoare pentru fiecare dintre ei. Daca \( g \) nu ar avea gradul mai mare decat 5, ar fi constant. Contradictie.
Yesterday is history,
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog