Ta có: A=ab+bc+ca
=10a+b+10b+c+10c+a
=(10a+10b+10c)+(a+b+c)
=10(a+b+c)+(a+b+c)
=11(a+b+c)\(⋮\)11
=>ĐPCM
\(A=\overline{ab}+\overline{bc}+\overline{ca}\)
\(\Rightarrow A=10a+b+10b+c+10c+a\)
\(\Rightarrow A=\left(10a+a\right)+\left(10b+b\right)+\left(10c+c\right)\)
\(\Rightarrow A=11a+11b+11c\)
\(\Rightarrow A=11\left(a+b+c\right)\)
Vì \(11⋮11\)
\(\Rightarrow11\left(a+b+c\right)⋮11\)
\(\Rightarrow A⋮11\left(đpcm\right)\)