E.511. Determinați ultimele 121212 cifre ale numărului n=52020−52019+2⋅52017+514⋅52016.n=5^{2020}-5^{2019}+ 2 \cdot 5^{2017} + 514 \cdot 5^{2016}.n=52020−52019+2⋅52017+514⋅52016.
Răspuns: 2500…0⏟10 de 0.25\underbrace{00 \ldots 0}_{\text{10 de 0}}.2510 de 000…0.
n=52016(54−53+2⋅5+514)=52016⋅1024=52006⋅(5⋅2)10.n=5^{2016}(5^4-5^3+2 \cdot 5 + 514) = 5^{2016} \cdot 1024 = 5^{2006} \cdot (5 \cdot 2)^{10}.n=52016(54−53+2⋅5+514)=52016⋅1024=52006⋅(5⋅2)10. Cum U2c(52006)=25⇒U12c(n)=2500…0⏟10 de 0.U_{2c}(5^{2006}) = 25 \Rightarrow \boxed{U_{12c}(n) = 25\underbrace{00 \ldots 0}_{\text{10 de 0}}}.U2c(52006)=25⇒U12c(n)=2510 de 000…0.