Question

Determinati restul impartirii la 5 a nr N = 2013 la puterea 2013 + 2014 la puterea 2014 + 2015 la puteres 2015

Answer 1

Uc(N)=Uc(2013^2013+2014^2014+2015^2015)=
=Uc[2013^(503×4+1)+2014^(1006×2+2)+5)=
=Uc(2013^1+2014^2+5)=
=Uc(3+6+5)=
=Uc(14)=4
restul va fi 4


explicatii
2013^1 se termina in 3
2013^2 se termina in 9
2013^3 se termina in 7
2013^4 se termina in 1
apoi se repeta
2013:4=503 si rest1
2013^2013 se termina ca 2013^1 adica in 3

2014^1 se termina in 4
2014^2 se termina in 6
apoi se repeta
2014:4=503 si rest2
2014^2014 se termina ca 2014^2 adica in 6

2015^2015 se termina in 5  deoarece 5  la orice putere se termina in 5