Question

determina ultima cifră a numărului 7 la puterea 2003 plus 8 la puterea 2003 plus 9 la puterea 2011

Answer 1

$u(7^{2003})= \\ \\ 2003:4=500,3 \\ \\ u(7^2^0^0^3)=u(7^3)=u(7*7*7)=u(3)$
$u(8^2^0^0^3)= \\ \\ 2003:4=500,3 \\ \\ u(8^2^0^0^3)=u(8^3)=u(8*8*8)=u(2)$
$u(9^2^0^1^1)= \\ \\ 2011:2=1005,1 \\ \\ u(9^2^0^1^1)=u(9^1)=u(9)$
$u(3)+u(2)+u(9)=u(4)$