Vì \(3n+4\in BC\left(5;n-1\right)\) nên \(\left(3n+4\right)⋮5\) và \(\left(3n+4\right)⋮n-1\)
Ta có:
\(\left(3n+4\right)⋮\left(n-1\right)\)
\(\Rightarrow\left\{\left(3n-3\right)+7\right\}⋮\left(n-1\right)\)
\(\Rightarrow\left\{\left(3n-1\right)+7\right\}⋮\left(n-1\right)\)
Vì \(3.\left(n-1\right)⋮\left(n-1\right)\) nên để \(\left\{3.\left(n-1\right)+7\right\}⋮\left(n-1\right)\) thì \(7⋮\left(n-1\right)\)
\(\Rightarrow n-1\in U\left(7\right)\)
\(\Rightarrow n-1\in\left\{1;7\right\}\)
\(\Rightarrow n\in\left\{2;8\right\}\)
Nếu \(n=2\) thì:
\(3n+4=3.2+4\)\(=6+4\)\(=10\)
Vì \(10⋮5\) nên \(n=2\) thỏa mãn.
Nếu \(n=8\) thì:
\(3n+4=3.8+4=24+4=28\)
Vì \(28⋮̸5\) nên \(n\ne8\)
Vậy \(n=2\)
