Question

Sa se determine ultimele trei cifre ale numarului n=2la 2009+4 la 1004+2la 2005

Answer 1

$2^{2009} + 2^{2*1004} + 2^{2005}= 2^{2009} + 2^{2008} + 2^{2005}= 2^{2005} ( 2^{4} + 2^{3} +1)= \\ =2^{2005}* (16+8+1)= 2^{2005} *25=2^{2005}* 5^{2} = 2^{2003} *100$
$u( 2^{1} )=2 \\ u( 2^{2} )=4 \\ u( 2^{3} )=8 \\ u( 2^{4} )=6 \\ u( 2^{5} )=2 \\ u( 2^{2003} )=8=>ultimele. 3 .cifre .sunt .800$

Answer 2

n=2²⁰⁰⁹+(2² )¹°°⁴+2²⁰⁰⁵
n=2²⁰⁰⁹+2²⁰⁰⁸+2²⁰⁰⁵
n=2²⁰⁰⁵(2⁴+2³+1)
n=2²°°⁵(16+8+1)
n=2²⁰⁰⁵×25
n=2²×2²⁰⁰³×5²
n=(2×5)²×2²⁰⁰³
n=10²×2²⁰⁰³
n=100×2²⁰⁰³
n=100×2³×2²⁰⁰⁰
n=100×8×2²⁰⁰⁰
n=800×2²⁰⁰⁰⇒ultimele trei cifre sunt.......800