Consider a smaller problem where the only digits are 0,1,2:

1001120221 would be such a number.  It has 10 digits.

There are 3^2 2-digit combinations of 0,1,2

So for A&B, such a number should have 10^2 + 1 or 101 digits.

I believe such numbers must exist using a graph theory argument:

Consider a graph with 10 nodes, one for each digit. Each node/digit connects to every other node with 2 pathways.  Each node also has one path connecting to itself.  Each node has an even number of connections.  So a circuit that covers every path exactly once ought to be possible.  (But I am uncertain if this is all that is necessary to prove it exists.)  

If true, a few observations follow:

- the path seems like it must start and stop at the same node

- each node will be touched 10 times except for the first/last node which is visited 11 times.

In the small example above, there are four 1s, three 0s, three 2s.  Also the number begins and ends with the same digit.