\(a\cdot10\cdot11+b\cdot11=1\cdot100+3\cdot10+b\)
\(a\cdot10+b\cdot10=3\cdot10\)
\(a+b=3=1+2=2+1\)
\(\Rightarrow\hept{\begin{cases}a=1\\b=2\end{cases}}\hept{\begin{cases}a=2\\b=1\end{cases}}\)
Ta có: ab. 11 = a3b
\(\Rightarrow\left(a.10+b\right).11=a.100+30+b\)
\(\Rightarrow a.110+b.11=a.100+b+30\)
\(\Rightarrow10a+10b=30\) ( bớt 2 vế cho 100a+b )
\(\Rightarrow10\left(a+b\right)=30\)
\(\Rightarrow a+b=30:10=3\)
\(\Rightarrow a+b=1+2=0+3\)
Mà a là chữ số hàng chục nên khác 0 suy ra \(ab\in\left\{30;21;12\right\}\)