👤

fie a=(1+2+3+...+2016) la puterea 1+2+3+...+2015 a) ultima cifra (a)=? b)aratati ca a e patrat perfect

Răspuns :

[tex]a)a=(1+2+3+..+2016)^{1+2+3+...+2015}\\ a=(\frac {2016*2017}2})^{\frac{2015*2016}{2}}\\ a=20133136^{2031120}\\ Toate\ puterile\ lui\ 6\ au\ ultime\ cifra\ 6\, deci\ ultima\ cifra\ este\ 6.\\ [/tex]
[tex]b)a=(1008*2017)^{2015*1008}\\ a=[(1008*2007)^{2015}]^{1008}\\ a=\{[(2023056)^{2015}]^{504}\}^{2}=\ \textgreater \ a\ este\ patrat\ perfect.[/tex]