\(n+S\left(n-3\right)=120\)
\(\Rightarrow n< 120\) (1)
\(\Rightarrow S\left(n-3\right)\le9+9=18\)
\(\Rightarrow n\ge120-18=102\) (2)
+ Từ (1) và (2) => n có dạng \(\overline{10x}\) hoặc \(\overline{11x}\)
+ TH1 : \(n=\overline{10x}\)
+ TH1.1 : n = 102
\(\Rightarrow n+S\left(n-3\right)=102+9+9=120\)( TM )
+ TH1.2 : \(n\ge103\Rightarrow S\left(n-3\right)=1+0+x-3\)
\(\Rightarrow n+S\left(n-3\right)=\overline{10x}+1+0+x-3=120\)
\(\Rightarrow100+x+x-2=120\)
\(\Rightarrow x=11\left(KTM\right)\) ( do x là chữ số )
+ TH2 : \(n=\overline{11x}\)
+ TH2.1 : n = 110
=> n + S(n-3) = 118 ( KTM )
+ TH2.2 : n = 111
=> n + S (n-3) = 111 + 1 + 8 =120 ( TM )
+ TH2.3 : n = 112
=> n + S (n-3) = 112 + 1 + 9 = 122 ( KTM )
+ TH2.4 : \(n\ge113\)
=> n + S (n-3) = \(\overline{11x}+1+1+x-3=120\)
\(\Rightarrow110-1+2x=120\)
\(\Rightarrow2x=11\Rightarrow x=\frac{11}{2}\left(KTM\right)\)
Vậy \(\left[{}\begin{matrix}n=102\\n=111\end{matrix}\right.\)