Question

$Sa~se~determine~primele~67~zecimale~ale~numarului~ \sqrt[3]{100...001}, \\ numarul~de~0-uri~e~de~2015.$

Answer 1

$\displaystyle Avem~1 \underbrace{00...0}_{\mbox{2015}}1=10^{2016}+1. \\ \\ \sqrt[3]{10^{2016}+1}\ \textgreater \ \sqrt[3]{10^{2016}}=10^{672}. \\ \\ Vom~demonstra~ca~primele~672~zeciamale~sunt~zerouri.~Pentru \\ \\ asta~este~suficient~sa~demonstram~ca~ \sqrt[3]{10^{2016}+1}\ \textless \ 10^{672}+ \frac{1}{10^{672}} . \\ \\ Ultima~relatie~se~verifica~prin~ridicare~la~cub: \\ \\ 10^{2016}+1\ \textless \ 10^{2016}+3 \cdot 10^{672}+ \frac{3}{10^{672}}+ \frac{1}{10^{2016}},~adevarat! \\ \\ (caci~3 \cdot 10^{672}\ \textgreater \ 1).$

$\displaystyle Rezulta,~deci,~ca~primele~672~zecimale~sunt~zerouri.$