Ta có:abcd-(ab+cd)=1000a+100b+10c+d-10a-b-10c-d=990a+99b=11(90a+9b)\(⋮11\)
Mà ab+cd\(⋮11\)\(\Rightarrow\)abcd\(⋮11\left(đpcm\right)\)
Ta có:
\(\overline{abcd}-\left(\overline{ab}+\overline{cd}\right)=100\overline{ab}+\overline{cd}-\overline{ab}-\overline{cd}=11.9\overline{ab}\)
Mà \(\overline{ab}+\overline{cd}\) và \(11.9\overline{ab}\) \(⋮\) 11 nên \(\overline{abcd}⋮11\)(đpcm)