Sa se rezolve in multimea numerelor naturale, ecuatia:
\( 3^{\left[\frac{x}{2}\right]+\left[\frac{x+1}{2}\right]+y-1}-11 \cdot 3^{x+1}=3888 \)
Cristian Calude, proba pe echipe, R.IV, P.III
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Raspuns
\(
\[
\begin{array}{l}
3^{[\frac{x}{2}] + [\frac{{x + 1}}{2}] + y - 1} - 11 \cdot 3^{x + 1} = 3888(*); \\
{\rm Sol:x = 2k} \Rightarrow {\rm 3}^{{\rm 2k + y - 1}} - 11 \cdot 3^{2k + 1} = 3888{\rm sau 3}^{{\rm 2k - 1}} (3^y - 99) = 3888; \\
{\rm Evident cautam y} \ge {\rm 5 si gasim ca doar perechea (4,5)convine;} \\
{\rm Daca x = 2k + 1 avem 3}^{{\rm 2k + y}} - 11 \cdot 3^{2k + 1} = 3888{\rm sau 3}^{{\rm 2k}} {\rm (3}^{\rm y} - 33) = 3888 \\
y \ge 4;{\rm si gasim ca doar (5,4)convine}{\rm .} \\
\end{array}
\]
\)
\[
\begin{array}{l}
3^{[\frac{x}{2}] + [\frac{{x + 1}}{2}] + y - 1} - 11 \cdot 3^{x + 1} = 3888(*); \\
{\rm Sol:x = 2k} \Rightarrow {\rm 3}^{{\rm 2k + y - 1}} - 11 \cdot 3^{2k + 1} = 3888{\rm sau 3}^{{\rm 2k - 1}} (3^y - 99) = 3888; \\
{\rm Evident cautam y} \ge {\rm 5 si gasim ca doar perechea (4,5)convine;} \\
{\rm Daca x = 2k + 1 avem 3}^{{\rm 2k + y}} - 11 \cdot 3^{2k + 1} = 3888{\rm sau 3}^{{\rm 2k}} {\rm (3}^{\rm y} - 33) = 3888 \\
y \ge 4;{\rm si gasim ca doar (5,4)convine}{\rm .} \\
\end{array}
\]
\)
La inceput a fost numarul. El este stapanul universului.