Question

va rog sa imi explicati cum se afla ultima cifra 1+3+3 la puterea2+3la puterea3+....3la puterea99

Answer 1

[tex]S=3^{0}+3^{1}+3^{2}+3^{3}+...+3^{99}|*3^{1}\\ 3S=3^{1}*(3^{0}+3^{1}+3^{2}+3^{3}+...+3^{99})\\ 3S=3^{1}*3^{0}+3^{1}*3^{1}+3^{1}*3^{2}+3^{1}*3^{3}+...+3^{1}*3^{99}\\ 3S=3^{1+0}+3^{1+1}+3^{1+2}+3^{1+3}+....+3^{1+99}\\ 3S=3^{1}+3^{2}+3^{3}+3^{4}+...+3^{100}\\ 3S=(3^{0}+3^{1}+3^{2}+3^{3}+...+3^{99})+3^{100}-3^{0}\\ 3S=S+3^{100}-3^{0}\\ 3S-S=3^{100}-1\\ 2S=3^{100}-1\\ S=\frac{3^{100}-1}{2}\\ 3^{1}=3\\ 3^{2}=9\\ 3^{3}=27\\ 3^{4}=81\\ 3^{5}=243\\ 3^{6}=729\\ Puterile\ lui\ 3\ se\ termina\ in:\ 3,9,7,1\\\frac{100}{4}=25\ rest\ 0\\U(3^{100})=U(3^{25*4})=U(3^{4})=U(81)=1\\U(3^{100}-1})=U(1-1)=0\\
U(\frac{3^{100}-1}{2}=U(\frac{0}{2})=0
[/tex]