Question

Determinați ultima cifră a numărului :
$a) \: x= {2}^{0} + {2}^{1} + {2}^{2} + {2}^{3} + ... + {2019}^{2019 } + {2020}^{2020} =$
$b) \: x = {2017}^{2017} + {2018}^{2018} + {2019}^{2019} + {2020}^{2020}$

DAU COROANĂ!!Explicație pas cu pas va rog !


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Answer 1

 

$\displaystyle\bf\\a)\\x=2^0+2^1+2^2+2^3+...+2^{2019}+2^{2020}=\boxed{\bf2^{2021}-1}\\\\U(2^{2021}-1)=U(2^{2020+1}-1)=U(2^{2020}\cdot2^1-1)=\\\\=U(2^{4\times 505}\cdot2^1-1)=U\Big(\Big(2^4\Big)^{505}\cdot2-1\Big)=U(16^{505}\cdot2-1)=\\\\=U(6^{505}\cdot2-1)=U(6\cdot2-1)=U(12-1)=U(11)=\boxed{\bf1}$

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$\displaystyle\bf\\b)\\\textbf{Vom calcula ultima cifra la fiecare putere apoi ultima cifra a sumei.}\\\\U\Big(2017^{2017}\Big)+U\Big(2018^{2018}\Big)+U\Big(2019^{2019}\Big)+U\Big(2020^{2020}\Big)=\\\\\\=U\Big(7^{2017}\Big)+U\Big(8^{2018}\Big)+U\Big(9^{2019}\Big)+U\Big(0^{2020}\Big)=\\\\\\=U\Big(7^{2016+1}\Big)+U\Big(8^{2016+2}\Big)+U\Big(9^{2018+1}\Big)+0=\\\\\\=U\Big(7^{4\times504+1}\Big)+U\Big(8^{4\times504+2}\Big)+U\Big(9^{2\times1009+1}\Big)+0=$

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$\displaystyle\bf\\=U\Big(\Big(7^4\Big)^{504}\times7^1\Big)+U\Big(\Big(8^4\Big)^{504}\times8^2}\Big) +U\Big(\Big(9^2\Big)^{1009}\times9^1}\Big)+0=\\\\\\=U\Big(2401^{504}\times7^1\Big)+U\Big(4096^{504}\times8^2}\Big) +U\Big(81^{1009}\times9^1}\Big)+0=\\\\\\=U\Big(1^{504}\times7^1\Big)+U\Big(6^{504}\times8^2}\Big) +U\Big(1^{1009}\times9^1}\Big)+0=\\\\\\=U\Big(1\times7\Big)+U\Big(6\times64}\Big) +U\Big(1\times9}\Big)+0=\\\\\\=7+U\Big(384}\Big) +9+0=U(7+4+9+0)=U(20)=\boxed{\bf0}$