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The Round Table

Answer

Here is the way of arranging the seven men:

A B C D E F G
A C D B G E F
A D B C F G E
A G B F E C D
A F C E G D B
A E D G F B C
A C E B G F D
A D G C F E B
A B F D E G C
A E F D C G B
A G E B D F C
A F G C B E D
A E B F C D G
A G C E D B F
A F D G B C E

Of course, at a circular table, A will be next to the man at the end of the line.

I first gave this problem for six persons on ten days, in the Daily Mail for the 13th and 16th October 1905, and it has since been discussed in various periodicals by mathematicians. 
Of course, it is easily seen that the maximum number of sittings for n persons is (n - 1)(n - 2)/2 ways. 
The comparatively easy method for solving all cases where n is a prime+1 was first discovered by Ernest Bergholt. 
I then pointed out the form and construction of a solution that I had obtained for 10 persons, from which E. D. Bewley found a general method for all even numbers. 
The odd numbers, however, are extremely difficult, and for a long time no progress could be made with their solution, the only numbers that could be worked being 7 (given above) and 5, 9, 17, and 33, these last four being all powers of 2+1. 
At last, however (though not without much difficulty), I discovered a subtle method for solving all cases, and have written out schedules for every number up to 25 inclusive. 
The case of 11 has been solved also by W. Nash. 
Perhaps the reader will like to try his hand at 13. 
He will find it an extraordinarily hard nut.





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