Question

Puteti va rog frumos sa ma ajutati?Dau coroana

Answer 1

 

$\displaystyle\bf\\U=ultima~cifra\\\\a)\\U\Big(3^{2019}+5^{2020}+1^{2021}\Big)=\\\\=U\Big(3^{2016+3}\Big)+U\Big(5^{2020}\Big)+U\Big(1^{2021}\Big)=\\\\=U\Big(3^{4\times504+3}\Big)+5+1=\\\\=U\left(\Big(3^4\Big)^{504}\times3^3}\right)+5+1=\\\\=U\left(\Big(81\Big)^{504}\times27}\right)+5+1=\\\\=U(1\times7+5+1)=U13=\boxed{\bf3}$

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$\displaystyle\bf\\b)\\U\Big(2^{2019}+6^{2020}2019^0\Big)=\\\\=U\Big(2^{2016+3}+6+1\Big)=\\\\=U\Big(2^{4\times504}\times2^3+6+1\Big)=\\\\=U\Big(16^{504}\times2^3+6+1\Big)=\\\\=U\Big(6\times8+6+1\Big)=\\\\=U\Big(48+6+1\Big)=\\\\=U\Big(15\Big)=\boxed{\bf5}$

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$\displaystyle\bf\\c)\\U\Big(4^{2019}+7^{2020}+8^{2021}-9^{2022}\Big)=\\\\=U\Big(4^{2018+1}+7^{2020}+8^{2020+1}-9^{2022}\Big)=\\\\=U\Big(4^{2\times1009}\times4^1+7^{4\times505}+8^{4\times505}\times8^1-9^{2\times1011}\Big)=\\\\=U\left(\Big(4^2\Big)^{1009}\times4^1+\Big(7^4\Big)^{505}+\Big(8^4\Big)^{505}\times8^1-\Big(9^2\Big)^{1011}\right)=$

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$\displaystyle\bf\\=U\Big(16^{1009}\times4^1+2401^{505}+4096^{505}\times8^1-81^{1011}\Big)=\\\\=U\Big(6^{1009}\times4+1^{505}+6^{505}\times8-1^{1011}\Big)=\\\\=U\Big(6\times4+1+6\times8-1\Big)=\\\\=U\Big(24+1+48-1\Big)=\\\\=U\Big(72\Big)=\boxed{\bf2}$