có $\overline{abcdeg}=\overline{ab} .10000+\overline{cd} .100+\overline{eg}=9999\overline{ab}+99\overline{cd}+(\overline{ab}+\overline{cd}+\overline{eg})=11.(909\overline{ab}+9\overline{cd})+(\overline{ab}+ \overline{cd}+ \overline{eg}) $$\Rightarrow \overline{abcdeg} $ chia hết cho $11$