Fermat's Little Theorem states that a^p-a=np where the variables are integers.
This can also be written a(a^(p-1)-1)=np so if a=10 we have 10(10^(p-1)-1)=np.
If p-1=3m where m is an integer then 10^m-1=10^((p-1)/3) is a factor of 10^3m-1.
X=(10^16-1)(10^10-1)=100(10^15-1)(10^9-1) and both exponents of 10 are multiples of 3.
So 3m=15 or 3m=9 making m=5 or 3. Therefore 10^5-1 and 10^3-1 are factors, that is, 99999 and 999.
999 is a listed choice, answer b.