Question

Tìm n nguyên dương để a= n³ + 31 chia hết cho n + 3

Answer 1

\(A=\dfrac{n^3+31}{n+3}=\dfrac{n^3+1+30}{n+3}=\dfrac{\left(n+3\right)\left(n^2-n+1\right)}{n+3}+\dfrac{30}{n+3}=n^2-n+1+\dfrac{30}{n+3}\)

n³ + 31 \(⋮\) n+3 \(\Leftrightarrow A\in Z\Leftrightarrow n^2-n+1+\dfrac{30}{n+3}\in Z\)

\(\Leftrightarrow n+3\inƯ\left(30\right)\)

\(\Leftrightarrow n+3=\left\{\pm1;\pm2;\pm3;\pm5;\pm6;\pm15;\pm30\right\}\)

mà n nguyên dương <=> \(n=\left\{2;3;12;27\right\}\)

Vậy...........................................