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Sa se determine cifrele x, y si z, stiind ca : \( \frac{1}{x+2y+3z}=\overline{0,xy} \).

\( \frac{1}{x+2y+3z}=\overline{0,xy}\Rightarrow \frac{1}{x+2y+3z}=\frac{\overline{xy}}{100}\Rightarrow \overline{xy}(x+2y+3z)=100\Rightarrow \overline{xy} \in \{10, 20, 25, 50\} \)
Analizand pentru fiecare valoare a lui \( \overline{xy} \), obtinem \( (x; y; z)\in \{(1; 0; 3); (2; 0; 1)\} \).

Andi Brojbeanu wrote:1/(x+2y+3z) =0,xy rezulta 100=(10x+y)(x+2y+3z). Cei doi factori apartin {1;2;4;5;10;20;25;50;100}. Analizand fiecare caz in parte, obtinem solutia finala (x;y;z) apartine {(1;0;3);(2;0;1)}.

\( \frac{1}{x+2y+3z}=\overline{0,xy} \Longrightarrow (10x+y)(x+2y+3z)=100 \Longrightarrow 10x+y\in \mathcal{D}_{100} \wedge \ x+2y+3z\in \mathcal{D}_{100} \). Analizand situatiile care apar obtinem ca \( \left(x,\ y,\ z\right)\in\left\{ \left(1,\ 0,\ 3\right),\ \left(2,\ 0,\ 1\right)\right\} \)