ABA and BAB

$\overline{ABA} =101A+10B =7d$

$\overline{BAB}=101B+10A$

$\overline{ABA} -\overline{BAB} =101(A-B)+10(B-A)=(A-B) \times 7 \times 13$ .

$\overline{BAB} =\overline{ABA} - (A-B) \times 7 \times 13 = 7(d-13(A-B))$

Note: The same condition holds for multiples of $13$ . It works for numbers $494, 585, 676, 767,858,949$ .

$\overline{ABA}=101A+10B$ is a multiple of $7$

thus $101A+10B-91A$ is also a multiple of $7$ $(91 A$ is a multiple of $7)$

$\Rightarrow 10A+10B$ is a multiple of $7$

$\Rightarrow 10A+10B+91B$ is a multiple of $7$ $(91 B$ is a multiple of $7)$

$\Rightarrow 101B+10A=100B+10A+B=\overline{BAB}$ is a multiple of $7$

For single (non-negative) digits $A,B$ and some positive integer $n$ we have that

$100A + 10B + A = 101A + 10B = 7n \Longrightarrow$

$3A + 3B = 7n - 98A - 7B = 7*(n - 14A - B) = 7m$ for some integer $m$ .

Now $\overline{BAB} = 100B + 10A + B = 7*(14B + A) + 3*(B + A) = 7k + 7m = 7d$

for some integer $d$ , and thus $\boxed{7 | \overline{ABA} \Longrightarrow 7 | \overline{BAB}}$ .

Comment: Note that $3(A + B) = 7m$ implies that $7|(A + B)$ , so the 3-digit number pairs $(\overline{ABA}, \overline{BAB}$ ) we are looking at are $(161,616), (252,525), (343,434), (595,959), (686,868)$ and of course $(777,777)$ .