Question

Arătați că n = 8^(33) +8^3 +2^(33) +2^3 se divide cu 82.

Answer 1

$Numarul~N=2^{99}+2^9+2^{33}+2^3~este~divizibil~cu~2,~deci~mai~ \\ \\ trebuie~sa~demonstram~ca~este~divizibil~cu~41. \\ \\ Avem:~N=2^9(2^{90}+1)+2^3(2^{30}+1). \\ \\ Vom~demonstra~ca~orice~numar~natural~de~forma~2^{10(2k+1)}+1, \\ \\ unde~k \in \mathbb{N},~este~divizibil~cu~41. \\ \\ 2^{10(2k+1)}+1=1024^{2k+1}+1=(1025-1)^{2k+1}+1 \\ \\ =M_{1025}+(-1)^{2k+1}+1=M_{1025}-1+1=M_{1025}~\vdots~41.$

$Prin~urmare~2^9(2^{90}+1)~si~2^{3}(2^{30}+1)~sunt~multipli~de~41,~si~ \\ \\ deci~N ~ \vdots~41,~iar~cum~N~ \vdots~2~si~(2,41)=1,~rezulta~ \boxed{N ~\vdots~82}~.$

$Am~folosit~proprietatea: \\ \\ (a+b)^n=M_a+b^n~ \forall~a,b,n \in .\mathbb{N^*}$