Concursul "Teodor Topan" - problema 1

[maky]

1. a) Aflati \( a \) daca \( \bar{0,1\left(a\right)}+\bar{0,a\left(3\right)}=0,3\left(5\right) \)
b) Se da \( n=2^{2006}-2^{2005}-2^{2004}-2^{2003} \). Aflati \( x \) din proportia: \( \frac{n}{x}=\frac{8^{667}}{5} \)
Crisan Georgeta, Simleul Silvaniei

[deleter]

\( 11a + 13 = 35 => 11a=22 => a=2 \)

[deleter]

11a+13=35 => 11a = 22 => a=2

[deleter]

\( n=2^{2005}*(2-1)-2^{2004}-2^{2003}=2^{2004}*(2-1) - 2^{2003} = 2^{2003}*(2-1)=2^{2003} \)
\( \frac{n}{x}=\frac{8^{667}}{5} => (2^{3})^{667}*x = 2^{2003}*5 => 2^{2001}x=2^{2003}*5 => 2^{2003}:2^{2001}=x:5 => 4=x:5 => x=20 \)