Question

Tìm n là số nguyên để A=1:(1/2011+1/2011+n) có giá trị nguyên.

Answer 1

Lời giải:

$A=\frac{2011(2011+n)}{4022+n}$

Để $A$ nguyên thì: $2011(2011+n)\vdots 4022+n$

$\Rightarrow 2011^2+2011(n+4022)-2011.4022\vdots 4022+n$

$\Rightarrow 2011^2-2011.4022\vdots 4022+n$

$\Rightarrow 2011^2-2011^2.2\vdots 4022+n$

$\Rightarrow 2011^2\vdots 4022+n$

$\Rightarrow 4022+n\in\left\{\pm 1; \pm 2011; \pm 2011^2\right\}$

$\Rightarrow n\in \left\{-4023; -4021; -2011; -6033; 4040099; -4048143\right\}$