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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++
\( \lg n!=\frac{n}{2}\Rightarrow n!=sqrt{10}^n\Rightarrow n=0 \)