Divide particpants over tables

[Jeroeny]

For a program I need to divide quiz participants over tables. Every table will have 1 winner who continues to the next round. Every table must have 6 to 9 participants.
so lets say you have 36 participants, you get 6 tables of 6 people, every table gets 1 winner and the finale has 1 table with 6 participants. But I need a formula so that I can enter a random number and get the amount of rounds and tables. I hope this is clear enough ( english isnt my native language )

 

[JeffM]

Question:
assume 54 participants
6 tables @ 9 participants each: leaves 6 winners for final table
9 tables @ 6 participants each: leaves 9 winners for final table

So if 54 is entered as random number, what happens?

There are lots of other similar cases...

Also:
minimum = 36 and maximum = 54, agree?

Off to the corner with you denis. 78 participants. First round, 8 tables of 9 each and 1 table of 6. Second round, 1 table of 9 winners from first round.

For certain numbers, the rules specified cannot be applied. An example is 11 participants. One table in the first round must have fewer than 6 participants.

Another example is 82 participants. In the first round, if there are 9 or fewer tables, then the average number per table is > 9, which violates the rules, but if there are 14 or more tables, then the average number per table < 6, which violates the rules. So the first round must consist of 10, 11, 12, or 13 tables, which will result in the same number of winners. But if there are only 10 or 11 winners, the second round must violate the rules. So the first round must consist of 12 or 13 tables, which will result in 12 or 13 winners. So the second round can proceed with 2 tables, but that results in just 2 winners for the third round, which violates the rules.

I have not proved this, but my intuition says that the rules can be applied strictly to p participants only if

\(\displaystyle \exists\ r \in \mathbb Z\ such\ that\ r > 0\ and\ 6^r \le p \le 9^r.\)

 

[Jeroeny]

Off to the corner with you denis. 78 participants. First round, 8 tables of 9 each and 1 table of 6. Second round, 1 table of 9 winners from first round.

For certain numbers, the rules specified cannot be applied. An example is 11 participants. One table in the first round must have fewer than 6 participants.

Another example is 82 participants. In the first round, if there are 9 or fewer tables, then the average number per table is > 9, which violates the rules, but if there are 14 or more tables, then the average number per table < 6, which violates the rules. So the first round must consist of 10, 11, 12, or 13 tables, which will result in the same number of winners. But if there are only 10 or 11 winners, the second round must violate the rules. So the first round must consist of 12 or 13 tables, which will result in 12 or 13 winners. So the second round can proceed with 2 tables, but that results in just 2 winners for the third round, which violates the rules.

I have not proved this, but my intuition says that the rules can be applied strictly to p participants only if

\(\displaystyle \exists\ r \in \mathbb Z\ such\ that\ r > 0\ and\ 6^r \le p \le 9^r.\)

That formula seems somewhat complicated, but I think i get it. Thanks!

 

[JeffM]

That formula seems somewhat complicated, but I think i get it. Thanks!

Well that formula does not give you the solution. I think, however, it specifies the numbers for which a solution is possible. I have not even proved the latter point.

 

[JeffM]

"To the corner I will go..."
But you tag along: 81, not 78.
9 tables of 9 participants: 9 weiners!
1 table with them 9 weiners...

No. No corner for me this time. Any number between 36 and 81 works. I was just using 78 as a counter-example to your maxing out at 54.

 

[JeffM]

No way, no way...I'm gonna tell my father on you!

I posted: minimum = 36 and maximum = 54, agree?

You answered: Off to the corner with you denis. 78 participants...
So you were answering my maximum question...

Come on over...we can play crazy eights...

OK Denis. I'm coming over, not because I need to, but for the entertainment