Not that $7|111111$, and we have $$\underbrace{66...66}_{2023} \star \underbrace{55...55}_{2023} \equiv 6 \star5 \underbrace{00...00}_{2022} \pmod {7}$$It's easy to figure out $\boxed{\star = 6}$ to make the large number divisible by $7$.

The first 2022 and the last 2022 digits are definitely divisible by 7, because 7 is a factor of 1001 and 1001 is a factor of both 666666 and 555555 Then all we need to make sure is that 605 + 10x = 0 mod 7 3x – 25 = 0 mod 7 3x = 4 mod 7 ===> x = 6 mod 7 Answer = 6