Question

1; Ultima cifra a nr 2006^n+2010^n+1+2005^n+2, n€N*, este......

2- Rezultatul calcului: (2+2•3)•{2•2^2•2^43-2^46+[3^148:3^43-(3^7)^15]+16}
Va rog este urgent! Dau coroana+ 20 pcte

Answer 1

1. N=2006^n+2010^n+1+2005^n+2
    ex. n=1
    u(2006^n)=6
    u(2010^n+1)=0
    u(2005^n+2)=5
    u(6+0+5)=1
 
    N*=numere nenule

         
2. (2+2x3){[2x2^2x 2^43-2^46+[3^148:3^43-(3^7)^15]+16}=
     =(2+6)[2^1x2^2x2^43-2^46+(3^105-3^105)+16]=
     =8(2^46-2^46+0+16)=
     =8x16=128
        

Answer 2

   
[tex]1) \\ Notatie:~~~U(x) = \texttt{Ultima cifra a lui x.} \\ U(2006^n+2010^{n+1}+2005^{n+2}) = \\ = U(6^n+0^{n+1}+5^{n+2})=U(6+0+5) =U(11)=\boxed{1} \\ \\ 2) \\ (2+2\cdot 3)\cdot \{2 \cdot 2^2 \cdot 2^{43}-2^{46}+[3^{148}:3^{43}-(3^7)^{15}]+16\} = \\ =(2+6)\cdot \{2^{1+2+43}-2^{46}+[3^{148-43}-3^{7\cdot 15}]+16\} = \\ =8\cdot \{2^{46}-2^{46}+[3^{105}-3^{105}]+16\} = \\ =8\cdot \{0+[0]+16\} = 8\cdot 16 = \boxed{128}[/tex]