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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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