Question

tìm số nguyên n để 2n+1 chia hết cho n-3

Answer 1

\(\hept{\begin{cases}^1\\ư\\a\end{cases}^2_z\overrightarrow{z}}\)

giúp mk!

Answer 2

\(2n+1⋮n-3\)

\(\Rightarrow2n-6+7⋮n-3\)

\(\Rightarrow2\left(n-3\right)+7⋮n-3\)

mà \(2\left(n-3\right)⋮n-3\Rightarrow n-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow n\in\left\{\pm4;10;2\right\}\)