Find the prime-power decomposition of 999999999999
As it was pointed in a commment $$999,999,999,999=10^{12}-1$$
Now, using difference of squares, we have $$10^{12}-1=(10^6-1)(10^6+1)=(10^3-1)(10^3+1)(10^6+1)$$
By sum/difference of cubes we have $$10^{12}-1=(10-1)(10^2+10+1)(10+1)(10^2-10+1)(10^2+1)(10^4-10^2+1)\\ =9 \cdot 111 \cdot 11\cdot 91 \cdot 101 \cdot 9901$$
Each of those numbers is easy to factor now.
P.S.
You can factor further $91$: $$91=100-9=10^2-3^2=(10-3)(10+3)$$
Hint:
$$999,999·1,000,001 = 999,999,999,999$$
$$999·1,001 = 999,999$$
You can figure out the rest ;D