Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\overline {abcd}  = \overline {ab00}  + \overline {cd}  = 100.\overline {ab}  + \overline {cd}  = 100.2\overline {cd}  + \overline {cd}  = 201\overline {cd} \\
201\,\, \vdots \,\,67 \Rightarrow 201\overline {cd} \,\, \vdots \,\,67 \Rightarrow \overline {abcd} \,\, \vdots \,\,67\\
2,\\
\overline {abc}  = 100a + 10b + c = \left( {2a + 3b + c} \right) + \left( {98a + 7b} \right)\\
 = \left( {2a + 3b + c} \right) + 7.\left( {14a + b} \right)\\
\left( {2a + b + c} \right)\,\, \vdots \,\,7\\
7\left( {14a + b} \right)\,\, \vdots \,\,7\\
 \Rightarrow \left[ {\left( {2a + 3b + c} \right) + 7.\left( {14a + b} \right)} \right]\,\, \vdots \,\,7\\
 \Rightarrow \overline {abc} \,\, \vdots \,\,7\\
3,\\
\left( {n + 5} \right)\,\, \vdots \,\,\left( {n - 2} \right)\\
 \Leftrightarrow \left[ {\left( {n - 2} \right) + 7} \right]\,\, \vdots \,\,\left( {n - 2} \right)\\
 \Leftrightarrow 7\,\, \vdots \,\,\left( {n - 2} \right)\\
 \Rightarrow \left( {n - 2} \right) \in Ư\left( 7 \right) = \left\{ {1;7} \right\}\\
 \Rightarrow n \in \left\{ {3;9} \right\}
\end{array}\)