Question

$Daca~ \frac{1}{10} + \frac{1}{11} + \frac{1}{12} +...+ \frac{1}{51}= \frac{a}{b} ~,a~si~b -naturale~prime~intre~ele. \\ Care~este~restul ~impartirii~lui~a~la~61 ~~? \\ (am~uitat~sa~mentionez~ca~a~si~b~sunt~si~nenule~)$

Answer 1

$S= \frac{1}{10} + \frac{1}{11} +...+ \frac{1}{51} = \\ \\ =( \frac{1}{10}+ \frac{1}{51})+( \frac{1}{11}+ \frac{1}{50})+...+( \frac{1}{25}+ \frac{1}{26})= \\ \\ = \frac{61}{10 \cdot 51}+ \frac{61}{11 \cdot 50}+...+ \frac{61}{25 \cdot 26} = \\ \\ =61(...) \\ \\ Deci~numaratorul~il~contine~pe~61~ca~factor,~in~timp~ce~ \\ \\ numitorul~nu~il~contine \Rightarrow Numarator ~ \vdots~61 \Rightarrow rest=0.$