>>101987891089 = 9.11^2

It is easy to check that the first operation yields a multiple of 9 and that reversing the digits preserves that condition, hence the second step also yields a multiple of 9

Secondly, the first step yields a multiple of 11 (same argument), and that given a multiple of 11, say with decimal expression abc (a+c=b), then abc+cba is a multiple of 121 that is at most 1089:

Write abc = a(11-1)^2 + b(11-1) + c.

Then abc = a*11^2 + (b-2a) 11

Similarly, cba = c*11^2 + (b-2c)11 and thus abc+cba = (a+c)11^2

Moreover, since a+c = b <= 9, we have abc+cba <= 1089.

Since the result of the process is divisible by 1089, it has to be equal to 1089.

Now would I have come up with that in 5 minutes during an interview, I'm not sure..