How do multi-round tourneys work?

So with these Correspondence/Multi-Day tournaments, how do multiple rounds work when the numbers don't mesh?

I would think that the numbers would need to mesh to be a legit tournament.  No?

For example:  If you had the setting 6(1) --> 2, then I would think you'd have to run either a 6-player, 18-player, or 54-player tournament.  With 54, you have 9 sections of 6, with in each case 2 advancing.  That's leaves 18 remain for round 2.  2 from each of the 3 sections would mean 6 for round 3.

But what I see a lot of is stuff like 34/96, 6(1) --> 2.  How on earth would this work?

You'd have 16 sections of 6 the first round.  Fine!

But then, if you have 2 from each section advance, that's 2 times 16, or 32.  32 doesn't divide by 6 evenly, so you end up with 5 sections of 6, and 2 odd men out.  WTF?

Same thing would happen with another I've seen, 50 players with a setting of 10(1) --> 3, and so 5 sections of 10, 3 advancing from each leaves you with 15 players, not a number divisible by 10.

Just doesn't make sense.  How would this ever work?

One would think ones like 12(1) --> 2 with 72 players, 12(1) --> 3 with 48 players, 12(1) --> 4 with 36 players, 10(1) --> 1 with 100 players, 10(1) --> 2 with 50 players, 8(1) --> 1 with 64 players, 8(1) --> 2 with 32 players, 6(1) --> 1 with 36 players, 6(1) --> 2 with either 18 or 54 players, 6(1) --> 3 with either 12, 24, 48, or 96 players, or many combinations with 4 or 2 per section would be the only ways that work.

bump

@OP: If you are referring to Chess.com Online Tournaments, then the groups are simply allowed to be uneven, as in this example.

If you are referring to true correspondence-chess tournaments not necessarily affiliated with Chess.com or other websites, then I cannot conclusively say how they are handled.

I hope that helps at least a little bit. If I don't understand you correctly, then I would be grateful for further clarification.

Was referring to Chess.com.  Thanks.

This leaves more questions for me.  Common sense tells me that the fairest solution to a potential spread in the group number for advanced rounds is to find the smallest spread proximate to that group number.  So for example, 6(1)-->2 with 42 players, there are 7 groups.  2 advance from each group.  Would  the advanced round include 2 groups of 7, or 2 groups of 5 and 1 group of 4? It would seem that 2 groups of 7 is a more equitable arrangement. If there is a spread between groups in the advanced round, how is it determined who is included in the larger/smaller group?  Is there any advantage given to the members in the group with the higher number of members, since they are given a bigger challenge to advance?