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Hint pentru o problema

Posted: Mon Mar 09, 2009 6:45 am
by alexandru_infomath
Am de rezolvat problema \( \lg1+\lg2+...+\lg n=n/2 \). (Pana acum am ajuns la \( \lg n!=n/2 \).)

Posted: Mon Mar 09, 2009 8:15 am
by mihai++
\( \lg n!=\frac{n}{2}\Rightarrow n!=sqrt{10}^n\Rightarrow n=0 \)