Question

Tìm n để

           a,   23n chia het cho n-1

           b,  \(n^2+n+2\) chia hết cho \(n+3\)

Answer 1

a) \(\frac{23n}{n-1}=\frac{23n-23+23}{n-1}=\frac{23\left(n-1\right)+23}{n-1}=23+\frac{23}{n-1}\)

\(\Rightarrow n-1\inƯ\left(23\right)\Rightarrow n-1\in\left\{-23;-1;1;23\right\}\Rightarrow n\in\left\{-22;0;2;24\right\}\)

b) \(\frac{n^2+n+2}{n+3}=\frac{n^2+3n-2n-6+8}{n+3}=\frac{n\left(n+3\right)-2\left(n+3\right)+8}{n+3}=n-2+\frac{8}{n+3}\)

\(\Rightarrow n+3\inƯ\left(8\right)\Rightarrow n+3\in\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

\(\Rightarrow n\in\left\{-11;-7;-5;-4;-2;-1;1;5\right\}\)

Answer 2

a) Ta có:

23n  chia hết cho n-1

=> 23n - 23 + 46 chia hết cho n - 1

=> 46 chia hết cho n - 1

=> n - 1 thuộc Ư(46) = {-1; 1; -2; 2; -23; 23; -46; 46}

=> n thuộc { 0; 2; -1; 3; -22; 24; -45; 47}

Vậy n thuộc { 0; 2; -1; 3; -22; 24; -45; 47}