Question

determonati nr nat n pentru care a=2^n+3^n+5^n se divide cu 5

Answer 1

[tex]u(2^{4k})=6;u(2^{4k+1})=2;u(2^{4k+2})=4;u(2^{4k+3})=8\\ u(3^{4k})=1;u(3^{4k+1})=3;u(3^{4k+2})=9;u(3^{4k+3})=7\\ u(5^{4k})=5;u(5^{4k+1})=5;u(5^{4k+2})=5;u(5^{4k+3})=5\\ u(2^{4k}+3^{4k}+5^{4k})=2\\ u(2^{4k+1}+3^{4k+1}+5^{4k+1})=0\\ u(2^{4k+2}+3^{4k+2}+5^{4k+2})=8\\ u(2^{4k+3}+3^{4k+3}+5^{4k+3})=0\\ Deducem\ ca\ a\vdots5\ daca\ n=4k+1\ sau\  n=4k+3,k\in N[/tex]