Question

Aflati ultima cifra a numerelor 3(la 2021)+4(la 2020)+7(la 2020)

Answer 1

 

$\displaystyle\\U\left(3^{2021}+4^{2020}+7^{2020}\right)=\\\\=U\left(3^{2021}\right)+U\left(4^{2020}\right)+U\left(7^{2020}\right)=\\\\=U\left(3^{2020+1}\right)+U\left(4^{2020}\right)+U\left(7^{2020}\right)=\\\\=U\left(3^{2020}\times3\right)+U\left(4^{2020}\right)+U\left(7^{2020}\right)=\\\\=U\left(3^{4\times505}\times3\right)+U\left(4^{2\times1010}\right)+U\left(7^{4\times505}\right)=\\\\=U\left(\Big(3^4\Big)^{505}\times3\right)+U\left(\Big(4^2\Big)^{1010}\right)+U\left(\Big(7^4\Big)^{505}\right)=$

$\displaystyle\\\\=U\left(81^{505}\times3\right)+U\left(16^{1010}\right)+U\left(2401^{505}\right)=\\\\=U\left(1^{505}\times3\right)+U\left(6^{1010}\right)+U\left(1^{505}\right)=\\\\=U\left(1\times3+6+1\right)=\\\\=U\left(3+6+1\right)=\\\\=U\left(10\right)=\boxed{0}$