These Solutions Look Cyclically Familiar

For any possible solution, we have $\begin{aligned} \overline{ABCDEF} & = \; 1000\overline{ABC} + \overline{DEF} \; = \; 999\big(1 + \overline{ABC}\big) \; = \; 3^3 \times 37 \big(\overline{ABC} + 1\big) \\ & = \; 10000\overline{AB} + 100\overline{CD} + \overline{EF} \; =\; 9999\overline{AB} + 99\overline{CD} + 99 \; = \; 99\big(101\overline{AB} + \overline{CD} + 1\big) \; =\; 3^2 \times 11\big(101\overline{AB} + \overline{CD} + 1\big) \end{aligned}$ and hence $3^3 \times 11 \times 37 = 10989$ divides $\overline{ABCDEF}$ . Since two particular solutions are $142857 \; = \; 13 \times 10989 \hspace{2cm} 241758 \; =\; 22 \times 10989$ we deduce that the greatest common divisor of all solutions is $\boxed{10989}$ .

In fact, there are $12$ solutions in total, all with the property that $A+D = B+E = C+F = 9$ . This means that, whenever $\overline{ABCDEF}$ is a solution, so is $\overline{CBAFED}$ .

A case, where $(B,D,F)$ and $(A,C,E)$ both add up to $9$ (satisfying the first condition), won't ever satisfy the second condition since the possible $\overline{ABC},\overline{DEF}$ won't be large enough to add up to $999$ .

So we should look for cases where $(B,D,F)$ add up to $19$ and $(A,C,E)$ add up to $8$ . $(B,D,F)=(4,7,8)$ and $(A,C,E)=(2,1,5)$ is a solution that satisfies both conditions. The corresponding integer $241758$ can be written as $2\times 3^3\times 11^2 \times 37$ . Another solution is $582417=3^3\times 11 \times 37 \times 53$ . Trying other possibilities hint that $3^3\times 11 \times 37$ exist in all possible solutions. In order to prove the hypothesis, we should notice the following

1- $1000\equiv 1$ when the modulo is $27,37$ . So $\overline{ABCDEF}$ can be written as $\overline{ABC}+\overline{ABC} \equiv 999 \equiv 0 \ mod \ 27\ ,\ 37$

2- the alternating sum of digits of $\overline{ABCDEF}$ is $19-8=11$ , meaning $\overline{ABCDEF}$ is divisible by $11$

Therefore, $27,37,11$ exist in all potential solutions (satisfying the two mentioned equations)