Question

Tìm số tự nhiên n sao cho n+3 chia hết cho n-1

Answer 1

\(\left(n+3\right)⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1\right)+4⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1\right)\inƯ\left(4\right)=\left\{-4;-2;-1;1;2;4\right\}\)

Do \(n\in N\)

\(\Rightarrow n\in\left\{0;2;3;5\right\}\)

Answer 2

(n+3)⋮(n−1)(n+3)⋮(n−1)

⇒(n−1)+4⋮(n−1)⇒(n−1)+4⋮(n−1)

⇒(n−1)∈Ư(4)={−4;−2;−1;1;2;4}⇒(n−1)∈Ư(4)={−4;−2;−1;1;2;4}

Vì