Page 1 of 1
Cristian Calude, proba pe echipe, R.III, P.I
Posted: Thu Nov 06, 2008 7:00 pm
by Laurian Filip
Sa se calculeze suma cifrelor numarului
\( N=9+99+999+...+\underbrace{999...9}_{999 cifre}. \)
Posted: Fri Jun 19, 2009 12:33 pm
by Andi Brojbeanu
\( N=(10-1)+(10^2)-1+(10^3-1)+....+(10^{999}-1)=10+10^2+10^3+....+10^{999}-999=10+10^2+10^4+....+10^{999}+10^3+1-10^3=10^{999}+....+10^4+10^2+10+1=111...10111 (1000 cifre)\Rightarrow \) Suma cifrelor numarului \( N \) este \( 999 \).
Posted: Fri Jun 19, 2009 12:45 pm
by spx2
\( N=(10-1)+(100-1)+...+(10^{999} - 1)=\underbrace{11...1}_{999 ori}0 - 999 =\underbrace{11...1}_{996 ori}0000 + 1110 - 999 = \underbrace{11...1}_{996 ori}0000 + 111 = \underbrace{11...1}_{996 ori}0111 \)
deci suma e 999