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test de selectie "seniori" Moldova 2014

Scris: Mar Feb 03, 2015 8:06 pm
de ghenghea1
1)Find all pairs of non-negative integers $(x,y)$ such that
$\sqrt{x+y}-\sqrt{x}-\sqrt{y}+2=0.$
2)Let $a,b\in\mathbb{R}_+$ such that $a+b=1$. Find the minimum value of the following expression:
$E(a,b)=3\sqrt{1+2a^2}+2\sqrt{40+9b^2}.$
3)Let $\triangle ABC$ be an acute triangle and AD the bisector of the angle $\angle BAC$ with $D\in(BC)$. Let E and F denote feet of perpendiculars from D to AB and AC respectively. If $BF\cap CE=K$ and $\odot AKE\cap BF=L$ prove that $DL\perp BF$.
4)Define $p(n)$to be th product of all non-zero digits of n. For instance $p(5)=5, p(27)=14, p(101)=1$ and so on. Find the greatest prime divisor of the following expression:
$p(1)+p(2)+p(3)+...+p(999)$.

Re: test de selectie "seniori" Moldova 2014

Scris: Mie Feb 04, 2015 9:40 am
de dangerous storm
Pentru 1:Avem $\sqrt{x}+\sqrt{y}-\sqrt{x+y}=2$,deci $2\sqrt{xy}=(\sqrt{x}+\sqrt{y}-\sqrt{x+y})(\sqrt{x}+\sqrt{y}+\sqrt{x+y})=$$2(\sqrt{x}+\sqrt{y}+\sqrt{x+y})$.Astfel obtinem $\sqrt{xy}=\sqrt{x}+\sqrt{y}+\sqrt{x+y}\le 2(\sqrt{x}+\sqrt{y})$.De aici se va observa ca $min\{\sqrt{x},\sqrt{y}\}\le 4$.Acum totul devine simplu,doar trebuie analizate niste cazuri.

Re: test de selectie "seniori" Moldova 2014

Scris: Mie Feb 04, 2015 1:06 pm
de DanDumitrescu
La problema 4 este o observatie foarte interesanta de facut si anume ca orice p(n) se scrie ca un produs de p(k) in functie de cate cifre are numarul.Vom considera p(o)=1 pentru a fi neutru in inmultire si vom avea Suma ciclica de p(n) de la 0 la 999 care va fi egala cu suma ciclica de p(n) n de la 0 la 9 ridicata la cub ce va genera 1000 de numere adica de la 0 la 999 iar suma ciclica de p(n) de la n=0 pana la n=9 va fi 46 si vom avea ca Suma ceruta si anume p(1)+p(2)+...+p(999)=46^3-1=45(46^2+46+1)=3*3*5*2163=3^3*5*7*103,deci rspunsul final este 103.

Re: test de selectie "seniori" Moldova 2014

Scris: Mie Feb 04, 2015 2:10 pm
de Marius Stănean
2) $\min=5\sqrt{11}$.