Ta có: \(\overline{abcabc}=\overline{abc}\cdot1000+\overline{abc}\)
\(=\overline{abc}\cdot\left(1000+1\right)=\overline{abc}\cdot1001\)
Vì \(1001⋮7\) nên \(\overline{abc}\cdot1001⋮7\)
hay \(\overline{abcabc}⋮7\).
abcabc = 1000.abc + abc
= 1001.abc
= 7.143.abc ⋮ 7
Vậy abcabc ⋮ 7