Question

Fie fractiile de forma $\frac{x+2016 }{2016x+1}$ , x apartine N. Determinati cel mai mic numar x pentru care fractia data se simplifica cu 2015.

Answer 1

$Pai~atat~numaratorul,~cat~si~numitorul~trebuie~sa~fie~multipli~de~ \\ \\ 2015.~Atunci~avem:~ \\ \\ x+2016=2015k~(k \in N,~k \geq 2) \Rightarrow x=2015k-2016. \\ \\ Numarul~2016x+1~trebuie~sa~fie~de~asemenea~un~multiplu~de~2015. \\ \\ 2016x+1=2015x+x+1=2015x+2015k-2016+1= \\ \\ =2015x+2015k-2015~ \vdots~2015~(a~iesit~de~la~sine~un~M_{2015}~:))~). \\ \\ Deci~fractia~se~simplifica~prin~2015,~pentru~x=2015k-2016 ~\forall~ \\ \\ k \in N,~k \geq 2.$

$k \geq 2 \Rightarrow 2015k-2016 \geq 2015 \cdot 2 -2016= \boxed{2014} \leftarrow~valoarea~minima.$