, 2019

For $a = 4$, an example of such a number is 126734895. For $a = 5$, an example is the number 549832761. (There are other solutions as well.)

We will show that for $a = 3, 6, 7, 8, 9$ there is no complete number with a difference number equal to $1a1a1a1a$. It then immediately follows that there is also no complete number $N$ with difference number equal to $a1a1a1a1$ (otherwise, we could write the digits of $N$ in reverse order and obtain a complete number with difference number $1a1a1a1a$).

For $a$ equal to 6, 7, 8, and 9, no such number $N$ exists for the following reason. For the digits 4, 5, and 6, there is no digit that differs by $a$ from that digit. Since the difference number of the complete number $N$ is equal to $1a1a1a1a$, every digit of $N$, except the first, must be next to a digit that differs from it by $a$. Hence, the digits 4, 5, and 6 can only occur in the first position of $N$, which is impossible.

For $a = 3$ the argument is different. If we consider the digits that differ by 3, we find the triples 1–4–7, 2–5–8, and 3–6–9. If the 1 is next to the 4 in $N$, the 7 cannot be next to the 4 and so the 7 must be the first digit of $N$. If the 1 is not next to the 4, the 1 must be the first digit of $N$. In the same way, either the 2 or the 8 must be the first digit of $N$ as well. This is impossible.