usually, the pattern for this is if the denominator is 9, 99, 999 or any other number equal to a power of 10 minus 1, gives you repeating digits.

3 also sometimes gives you repeating digits since is a factor of 9.

in this problem, 2 and 5 are out b/c they are terminating.

12 and 15 are out as well too b/c 8103/12 = 2701/4 and 8103/15 = 2701/5 --> both numerator and denominator are divisible by 3.

4 and 5 are terminating denominators. Only option left is 9.