Ta có:
\(\overline{ab}+\overline{ba}=10a+b+10b+a=\left(10a+a\right)+\left(10b+b\right)=11a+11b=11\left(a+b\right)⋮11\)
\(\Rightarrow\overline{ab}+\overline{ba}⋮11\left(đpcm\right)\)
Ta có ab+ba=a.10+b.1+b.10+a.1
= (a.10+a.1)+(b.1+b.10)
= a.(10+1)+b.(10+1)
= a.11+b.11
Vì \(11⋮11\)
\(\Rightarrow a.11+b.11⋮11\)
hay ab+ba chia hết cho 11