Question

Tìm số tự nhiên n để 3n+29 chia hết cho n+3

Answer 1

\(\dfrac{3n+29}{n+3}=\dfrac{3\left(n+3\right)+20}{n+3}=3+\dfrac{20}{n+3}\)

Để \(3n+29⋮n+3\Rightarrow20⋮n+3\)

Hay n+3 là ước của 20 do n là số tự nhiên \(\Rightarrow\left(n+3\right)\ge3\)

\(\Rightarrow\left(n+3\right)=\left\{4;5;10;20\right\}\Rightarrow n=\left\{1;2;7;17\right\}\)

 

 

Answer 2

\(3n+29⋮n+3\)

\(\Rightarrow3n+29-3\left(n+3\right)⋮n+3\)

\(\Rightarrow3n+29-3n-9⋮n+3\)

\(\Rightarrow20⋮n+3\)

\(\Rightarrow n+3\in\left\{-1;1;-2;2;-4;4;-5;5;-20;20\right\}\)

\(\Rightarrow n\in\left\{-4;-2;-5;-1;-7;1;-8;2;-23;17\right\}\left(n\in Z\right)\)