Ta có : \(\overline{abcd}=10\overline{ab}+\overline{cd}=100.2.\overline{cd}+\overline{cd}\)
\(=201.\overline{cd}\)
Mà \(201⋮67\)nên \(201.\overline{cd}⋮67\)
Vậy \(\overline{abcd}⋮67\)
Ta có: abcd = ab x 100 + cd =200cd +cd (vì ab = 2cd)
hay=201cd
Mà \(201⋮67\left(=3\right)\)
\(\Rightarrow201\overline{cd}⋮67\)
Vậy \(\overline{ab}=2\overline{cd}\Leftrightarrow\overline{abcd}⋮67\)