Question

Determinati ultimele trei cifre ale numarului n=2la puterea2006 -2 la puterea 2004 +2 la puterea 2001

Answer 1

$n=2^{2006}-2^{2004}+2^{2001}$

$n=2^{2001} (2^{5} -2^{3} +1)$

$n=2^{2} x2^{1999}x25$

$n=100x2^{1999}$

$n=100x2^{1996+3}$

$u(2^{1999})=u(2^{3})$

$u(2^{1999})=8$

ultimele 3 cifre ale nr n sunt: 8·100=800