Question

determinați ultima cifră a numărului n, unde n=1+1+2+2^2+2^3+...+2^2018​

Answer 1

n=1+2¹+2²+2³+...+2²⁰¹⁸

2n=2¹+2²+2³+....+2²⁰¹⁸+2²⁰¹⁹

-----------------------------------(-)

n=2²⁰¹⁹-1

u(2¹)=2

u(2²)=4

u(2³)=8

u(2⁴)=6

u(2⁵)=2

2019/4=504 rest 3 => u(2²⁰¹⁹)=u(2³)=8

u(n)=u(2²⁰¹⁹)-1=u(2³)-1=8-1=7

Deci u(n)=7