Có: \(\overline{20a20a20a}=200200200+\overline{a00a00a}=200200200+1001001\cdot a\)
Ta thấy: \(\left\{{}\begin{matrix}200200200\equiv4\left(mod7\right)\\1001001\equiv1\left(mod7\right)\end{matrix}\right.\)
Nên \(\overline{20a20a20a}⋮7\) khi \(1001001\cdot a\equiv3\left(mod7\right)\)
Vậy ta được a = 3