"Marian Tarina", 2006

Claudiu Mindrila · Post by **Claudiu Mindrila** » Wed Apr 23, 2008 1:34 pm

Sa se determine \( n \in \mathbb{N}^* \) astfel incat \( \frac{{2\sqrt n - 5\sqrt 3 }}{{\sqrt 3 + \sqrt n }} \in \mathbb{Z} \).

Concursul Marian Tarina, 2006, Mariana Ursu si Gheorghe Lobont

Ahiles · Post by **Ahiles** » Wed Apr 30, 2008 1:02 pm

Problema foarte cunoscuta.... Asemenea problema cu radicali a fost la Olimpiada Municipala Chisinau in clasa 7.
\( \frac{2\sqrt{n}-5\sqrt{3}}{\sqrt{3}+\sqrt{n}}=\frac{2(\sqrt{n}+\sqrt{3})-7\sqrt{3}}{\sqrt{3}+\sqrt{n}}= 2-\frac{7\sqrt{3}}{\sqrt{3}+\sqrt{n}}. \)
Deci \( \frac{7\sqrt{3}}{\sqrt{3}+\sqrt{n}}\in \mathbb{Z} \).
De unde \( \sqrt{3}+\sqrt{n}\in \{\sqrt{3} ; 7\sqrt{3}} \), deci \( \sqrt{n}\in \{0;6\sqrt{3}\} \) sau \( n\in\{0; 108} \).

Virgil Nicula · Post by **Virgil Nicula** » Sat May 17, 2008 9:00 am

Ahiles wrote:Deci \( \frac{7\sqrt{3}}{\sqrt{3}+\sqrt{n}}\in \mathbb{Z} \). De unde \( \sqrt{3}+\sqrt{n}\in \{\sqrt{3} ; 7\sqrt{3}} \) .

DE CE ?!

Incearca n oricare dintre 12 , 48 , 300 , 30000, 243, etc si ai sa vezi ca merge.

Asa ca degeaba este ... foarte cunoscuta.

Claudiu Mindrila · Post by **Claudiu Mindrila** » Mon May 19, 2008 6:43 pm

Indicatie: Amplificati fractia ce trebuie sa apartina \( \mathbb{Z} \) cu \( \sqrt 3-sqrt n \)