Question

3n+5 chia hết cho n-3

Answer 1

    \(3n+5\) ⋮ \(n-3\)
⇔\(\left(n-3\right)+\left(n-3\right)+\left(n-3\right)+16\) ⋮ \(n-3\)
Để \(3n+5\) ⋮ \(n-3\)
thì \(n-3\) ∈ \(Ư\left(16\right)\)
    \(Ư\left(16\right)=\left\{\pm1;\pm2;\pm4;\pm8;\pm16\right\}\)
⇒ \(n-3=\left\{\pm1;\pm2;\pm4;\pm8;\pm16\right\}\)
⇒ \(n=\left\{4;2;5-1;7;-1;11;-5;19;-13\right\}\)

\(\#PeaGea\)

Answer 2

 Ta có : 3n + 5 ⋮ n - 3

=> (3n - 9) + 14 ⋮ n - 3

=> 3(n - 3) + 14 ⋮ n - 3

Vì 3(n - 3) ⋮ n - 3 nên 14 ⋮ n - 3

=> n - 3 ∈ Ư(14) ∈ {-14;-7;-2;-1;1;2;7;14}

=> n ∈ {-11;-4;1;2;4;5;10;17}