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Divizibilitate
Posted:
Sat Jan 23, 2010 7:52 pm
by
Mateescu Constantin
Demonstrati ca
\( \(2^{2x}+2^{x+y}+2^{2y}\)!\ \vdots\ \(2^x!\)^{2^x+2^{y-1}}\ \cdot\ \(2^y!\)^{2^y+2^{x-1}}\ ,\ \forall\ x,y\in\mathbb{N}^{\ast} \)
.