Question

Determinati ultima cifra a numarului:
a) 30^71+31^71+35^71.
b)6+6^2+6^3+6^4+.....+6^40.
c)2^1999•5^2000+3.
d)2^1993•6^1890•7^1234•5^1993•9•1^1564.
e)3^1986+4^1987+5^1988.
Vă rog! Dau coroană!!!REPEDE!

Answer 1

$a)\ U(30^{71}+31^{71}+35^{71})=U(0+1+5)=6\\U(30^{71})=0 (Multipli\ de\ 10\ la\ orice\ putere\ o\ sa\ aiba\ ultima\ cifra\ 0)\\U(31^{1})=1\\U(31^{2})=1\\Atunci\ U{31^{71}}=1\\U(35^{1})=5\\U(35^{2})=5\\Atunci U(35^{71})=5$

Observatie:

Orice numar care se termina in 0 va  avea, ridicat la orice putere,ultima cifra 0. Orice numar care se termina in 5 va avea, ridicat la orice putere, ultima cifra 5. Orice numar care se termina in 1 va avea, ridicat la orice putere, ultima cifra 1. Orice numar care se termina in 6 va avea, ridicat la orice putere, ultima cifra 6.

$b)\ U(6+6^{2}+....+6^{40})=U(6+6+...+6)=U(6*40)=U(240)=0\\c)\ U(2^{1999}*5^{2000}+3)=U(8*5+3)=U(43)=3\\U(2^{1})=2\\U(2^{2})=4\\U(2^{3})=8\\U(2^{4})=6\\U(2^{5})=2\\Inseamna\ ca\ cifra\ lui\ doi\ se\ repeta\ din\ 4\ in\ 4.\\1999:4=499\ rest\ 3\ ->\ U(2^{1999})=8$

$d)\ U(2^{1993}*6^{1890}*7^{1234}*5^{1993}*9*1^{1564})=U(2*6*9*5*9*1)=U(4860)=0\\1993:4=498 rest\ 1\ ->/ U(2^(1993))=2\\U(7^{1})=7\\U(7^{2})=9\\U(7^{3})=3\\U(7^{4})=1\\U(7^{5})=7\\Inseamna\ ca\ ultima\ cifra\ a\ lui\ sapte\ se\ repeta\ din 4\ in\ 4.\\1234:4=308\ rest\ 2\ ->\ U(7^{1234})=9$

$e)\ U(3^{1986}+4^{1987}+5^{1988})=U(9+4+5)=U(18)=8\\U(3^{1})=3\\U(3^{2})=9\\U(3^{3})=7\\U(3^{4})=1\\U(3^{5})=3\\Inseamna\ ca\ ultima\ cifra\ a\ lui\ 3\ se\ repeta\ din\ 4\ in\ 4.\\1986:4=496\ rest\ 2/ ->/ U(3^(1986))=9\\U(4^{1})=4\\U(4^{2})=6\\U{4^{3}}=4\\Inseamna\ ca\ ultima\ cifra\ a\ lui\ 4\ se\ repeta\ din 2\ in\ 2.\\ 1987:2=993\ rest\ 1\ ->\ U(4^{1987})=4$