The remainder, when a number n is divided by 6, is p. The remainder

GMATNinja KarishmaB Bunuel: Any quicker way to solve this problem? I took 3:32 mins!!

Okay:
One way would be to begin with statement 2, because "p" will always be 0. But I still have to do the 6 multiplication table and find the remainder pattern for n/12. (will take about a min)
In statement 1, I would still have to find the remainder pattern n = 8K (8k/6 and 8k/12) before I know statement 2 alone is insufficient.
I'll combine both statements then. It would still take me like 3 minutes to answer this question. How do I reduce the time?

Check this video on division and remainders first: https://youtu.be/A5abKfUBFSc

The remainder, when a number n is divided by 6, is p. The remainder, when the same number n is divided by 12, is q.
This tells me that EITHER q = p OR q = p + 6 (so p < q) depending on how many complete groups of 6 there are in n. If n has odd number of groups of 6, then p < q.
If n has even number of groups of 6, then p = q. So we need to find whether n has odd number of groups of 6 or even number of groups of 6 to find whether p < q or not.

(1) n is a positive number having 8 as a factor.

Here n can have odd number of groups of 6 (when n = 8) or even (when n = 16). Not sufficient.

(2) n is a positive number having 6 as a factor.

Again n can have odd number of groups of 6 (when n = 6) or even (when n = 12). Not sufficient.

Using both, we see that n has 8 and 6 as factors so it certainly has 24 as factor (LCM of 6 and 8). This means that n has an even number of groups of 6 and hence q = p
or simply, n is divisible by 24 means it is divisible by both 6 and 12 hence p = q = 0. Hence answer is "definite No".

Answer (C)