Question

Va rog ajutatima....5<6<7....URGENT va rog

Answer 1

       
$5) \\ \text{ Sa se afle ultimele 2014 cifre ale numarului: } 7000^{671}. \\ 7000^{671} = (7\cdot1000)^{671}=7^{671}\cdot1000^{671} \\ 1000^{671} = 1 \;urmat \;de\; 3\cdot671\; zerouri =1 \;urmat \;de\; 2013\; zerouri \\ U = ultima cifra. \\ U(7^{671})=U(7^{668+3})=U(7^{668})\cdot U(7^{3})=U(7^{4 \cdot 167})\cdot U(7^{3})= \\ =U((7^4)^{167}) \cdot U(7^{3})=U(2401^{167}) \cdot U(343)=1 \cdot 3 = 3 \\ =\ \textgreater \ \;\; \text{Ultimele 2014 cifre ale numarului: }7000^{671} \;sunt: \\ \boxed{3000...de \;2013 \;ori...000}$


$6) \\ Aflati\; ultima \;cifra \;a \;numarului \; \;\;a = 2^{2013} \cdot 5^{2014} + 7^5 .\\ \\ U( 2^{2013}) = U( 2^{2012+1}) =U(2\cdot 2^{2012})=U(2\cdot 2^{4 \cdot 503 })= \\ =U(2\cdot 16^{503})=U(2 \cdot 6)=\boxed{2} \\ \\ U(5^{2014})=\boxed{5} \\ \\ U(7^5) = U(7^{4+1})=U(7 \cdot7^4)=U(7 \cdot7^{2\cdot2})=U(7 \cdot49^{2})= 7\cdot 1=\boxed{7} \\ \\ =\ \textgreater \ \;\;\;U(2^{2013} \cdot 5^{2014} + 7^5)=U(2 \cdot 5+7)=U(17) =\boxed{\boxed{7}}$


$7) \\ Aratati \;ca \;numarul\; 25^{2013} \;este \;patrat \;perfect. \\ Rezolvare: \\ 25^{2013}=(5\cdot5)^{2013} = 5^{2013}\cdot5^{2013}= (5^{2013})^2 \;\;\;\;=\ \textgreater \ \;\;\; p.p.$