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Manabeatricebea
a fost răspuns

Calculati ultima cifara a numerelor:
A)2 pe 47,2 pe129, 2 pe2000
B)21 pe324,19 pe257,17 pe2000
C)5 pe1999+7 pe 2000;
D)3 pe1992+4 pe 1994+5 pe1996+6 pe1998+7 pe 2000

Răspuns :

Raducustoica1
U(2^47) = U(2^7)= ...8
U(2^129)= U(2^1)=2
U(2^2000)=U(2^0)=1
U(21^324)=U(1^0)=1
U(19^257)=U(9^1)=9
U(17^2000) = U(7^0) = 1
U(5^1999) + U(7^2000) = U(5^3)+U(7^0)= ...5 +1=...6
U(3^1992) + U(4^1994)+U(5^1996) + U(6^1998)+ U(7^2000)=
U(3^0)+U(4^0)+U(5^0)+U(6^2)+U(7^0)=1+1+1+..6+1=..0
Ilaura15
U(2^47) = U(2^7)= ...8
U(2^129)= U(2^1)=2
U(2^2000)=U(2^0)=1
U(21^324)=U(1^0)=1
U(19^257)=U(9^1)=9
U(17^2000) = U(7^0) = 1
U(5^1999) + U(7^2000) = U(5^3)+U(7^0)= ...5 +1=...6
U(3^1992) + U(4^1994)+U(5^1996) + U(6^1998)+ U(7^2000)=
U(3^0)+U(4^0)+U(5^0)+U(6^2)+U(7^0)=1+1+1+..6+1=..0

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