Đáp án:
$n \in \left\{ {4;\,\,6;\,\,10;\,\,24} \right\}.$
Giải thích các bước giải:
Ta có: \(n - 3\) chia hết cho \(n - 3\)
\(\begin{array}{l} \Rightarrow 7\left( {n - 3} \right)\,\, \vdots \,\,\left( {n - 3} \right)\\ \Rightarrow 7n - 21\,\,\, \vdots \,\,\,n - 3\end{array}\)
Mà \(7n\,\, \vdots \,\,\,n - 3 \Rightarrow 21\,\,\, \vdots \,\,n - 3\)
\( \Rightarrow n - 3 \in U\left( {21} \right)\)
Lại có: \(U\left( {21} \right) = \left\{ {1;\,\,3;\,\,7;\,\,21} \right\}\)
\(\begin{array}{l} \Rightarrow \left[ \begin{array}{l}n - 3 = 1\\n - 3 = 3\\n - 3 = 7\\n - 3 = 21\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}n = 4\\n = 6\\n = 10\\n = 24\end{array} \right..\\ \Rightarrow n \in \left\{ {4;\,\,6;\,\,10;\,\,24} \right\}.\end{array}\)
