The Dinner and Dialogue Club has planned a series of small meetings. Each meeting would consist of two or three members enjoying friendly conversation with each other while eating food from different places all over the world. Each member is scheduled to meet exactly four times. No pair of members will meet twice, but some pairs might not meet at all.
The first thing the club did was schedule and arrange the meetings so that each member knew whom to meet and when. When it came to choosing restaurants, someone suggested that each member eat at two restaurants with eastern food, and two with western food (each restaurant is either one or the other). They liked the idea, but to their dismay, the idea was not possible without rearranging at least some of the meetings.
What possible meeting schedule might cause this to happen? How many members are there in this club, at the least?
We could let:
n = number of members;
x = meetings of 2; y = meetings of 3;
Then 2x+3y = 4n and this has an infinite number of non-negative integer solutions: x = a, y = 2a+4b , n= 2a+3b for {a,b} integers with.a >= 0; b >= -a/2.
Example: a=10, b=-5 shows a 10 x 2-meeting solution for n = 5 is possible or more strictly is not impossible.
At each meeting either two or three meals is eaten.
So if we let x = e+f and y=g+h, then the number of eastern meals is 2e+3g and western meals 2f+3g. If, we require the meeting solution to be true but 2e+3g notEqual 2f+3g for all e,f,g,h>=0, then we may have a solution.
It is looking BIG at the moment.
|
|
Posted by goFish
on 2005-11-04 06:38:31 |
First thoughts
