Salut,
[tex]x_{n+1}=(10-1)^{x_n}=(-1+10)^{x_n}=-C_{x_n}^0+C_{x_n}^1\cdot 10 +M_{100}=\\\\=M_{100}+10x_n-1,\ unde\ M_{100}\ este\ multiplu\ de\ 100.[/tex]
[tex]Deci\ U2C(x_{n+1})=U2C(10x_n-1),\ unde\ U2C-\ ultimele\ dou\breve{a}\ cifre.\\\\Avem\ c\breve{a}\ U2C(x_2)=U2C(10\cdot 9-1)=89,\ iar\ U2C(x_3)=u(10\cdot 89-1)=89.\\\\Mai\ departe\ se\ poate\ ar\breve{a}ta\ la\ fel\ c\breve{a}\ ultimele\ 2\ cifre\ sunt\ tot\ 89.\\\\\c{S}irul\ din\ enun\c{t}\ are\ forma\ X=10n+9\ (ultima\ cifr\breve{a}\ a\ lui\ X\ este\ 9),\ deci:\\\\U2C(10X-1)=U2C[10(10n+9)-1]=U2C(100n+89)=89.[/tex]
Deci ultimele 2 cifre sunt 89.
Green eyes.
