Divizibilitate

alex2008 · Post by **alex2008** » Wed Jan 14, 2009 9:34 am

Fara a face efectiv calculele demonstrati ca numarul \( 2008^{10}+2008^5+1 \) este divizibil cu numarul \( 2008^2+2009 \) .

Etapa locala , Bihor 2008 , Romulus Plesa si Viorica Plesa

DrAGos Calinescu · Post by **DrAGos Calinescu** » Wed Jan 14, 2009 11:11 pm

\( 2008^{10}+2008^5+1=2008^{10}+2008^9+2008^8-2008^8-2008^7-2008^6+2008^7+2008^6+2008^5-2008^6-2008^5-2008^4+2008^5+2008^4+2008^3-2008^3-2008^2-2008
+2008^2+2008+1 \)

In grupe de cate trei se da factor comun si iese fiecare multiplu de \( 2008^2+2009 \)