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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

I know. I think that she was a moron but she was right about me too.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**TracyBidwell****Member**- Registered: 2013-12-18
- Posts: 3

ShivamS wrote:

It's likely due to the fact that when you google "is 0 a rational number?", the first creditable result is from MIF.

Yup! Now, nothing is possible you can find the answer immediately in a blink of an eye. So if you have any fact to find the first thing to do is search in google.

Habeeb Akande

If you are not working towards something, your life will end with nothing.

― Habeeb Akande

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 8

There are 4 progressive dinners, 5 groups of 4 couples each (20 couples alltogether). Each dinner has 4 courses. Using each couple as a number from 1 to 20, how can it work out so that each couple (number) does only 1 course and is never with any other couple more than once?

*Last edited by allan1085 (2013-12-19 10:11:38)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

As far as I know 5 groups of 4 couples as you describe is not possible. This is the closest answer.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 8

Hi bobbym, I guess I didn't explain myself very well. There are only 4 dinners. 5 groups of 4 each. The numbers must be scrambled so that each course is by a different couple. For example, you have #1 serving the 1st course each dinner.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

I probably can permute those first numbers, but if days = dinners then it is impossible to have 4.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 8

bobbyum, Last year I was able to work it out for 16 couples, 4 to a group, 4 dinners. This year we added 4 more couples, but still only 4 dinners having 5 groups for each dinner. Yes, days=dinners.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Not everyone of these progressive dinners has a solution, most do not. I am looking at the charts for all known solutions. 16 couples I believe is possible. I am seeing no solution for 20 couples, 4 dinners, 5 groups.

Now all of these progressive dinners have the rule that each couple eats with every other couple once and only once. If you want to relax the criterion and have the couples not eat with every other one then more solutions are possible.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 8

bobbym, Yes, your suggestion is fine. Any solutions you can give me would be gresatly appreciated.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

Since you require 4 dinners that means each couple can see 3 new couples per dinner. 4 x 3 = 12 couples maximum for the 4 dinners. That means each couple ideally will miss eating with 7 couples.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 8

bobbym, No, there are still 20 couples. However, if some of the couples see each other more than once, than so be it. They just can't serve the same course more than once. I really appreciate ypour help.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Look at couple #1. Take the first 4 dinners 1, 2, 3, 4.

1 eats with 2,3,4,5,6,9,10,13,15,16,18,20

Notice that 7,8,11,12,14,19 and 17 are missing.

1 ( and everybody else too ) will always be missing at least 7 people. This is because each dinner he eats with 3 different people. 4 x 3 = 12. 19 - 12 = 7.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 8

bobym, The problem is that you have #1 serving the 1st course of 5 dinners. I am trying to get it so that no couple serves the same course more than once. If some couples are together more than once, than so be it. They just can't serve the same course more than once.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

I am working on that, I will post an answer as soon as I get it.

The solution may take awhile.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 8

thanki you

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

How does this look?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**allan1085****Member**- Registered: 2013-12-19
- Posts: 8

thank you bobbym, you have been a great help.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Hi;

Let me know if you need any adjustments. I did it by hand so there may still be inconsistencies in the arrangements.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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