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 Subject: RE: Ten Weary Travelers and 1-1 correspondence Author: Steve Date: Apr 11 2004
Hi Anon,
I think that you are too clever for this riddle.  A more un-trained eye
(such as me when I first read through this) would miss the fact that one of the
people is being double-counted.  The first and second traveler are placed in
room A.  At the end, He re-labels one of these two travelers as #10 and places
him in room I, which leaves out the real 10th traveler.
The only other thing I can think of is that he could have placed the 10th
traveler in I before lodging anyone else, so that the 10th traveler got his
sleep WHILE the landlord was placing everyone else in their rooms.  Then the
first or second traveler would take the 10th traveler's place in I.

On Apr 10, 2004, Anon wrote:

I am trying to figure this out and am totally lost.

Ten Weary, footsore travelers, all in a woeful plight, sought shelter at a
wayside inn one dark and stormy night.

Nine rooms, no more the landlord said, Have I to offer you.  To each of you a
single bed, but the ninth must serve for two.

A din arose. The troubled host could only scratch his head, for those tired men
no two would occupy one bed.

The puzzled host was soon at ease- he was a clever man- and so to please his
guests devised this most ingenious plan.

In room marked A, two men were placed, the third was lodged in B, the fourth to
C was then assigned, the fifth retired in D.

In E the sixth he turned away, in F the seventh man, the eighth and ninth in G
and H, and then to A he ran,

Wherein the host as I have said, had laid two travelers by, then taking
one-the tenth and last- he lodged him safe in I.

Nine single rooms-a room for each-were made to serve for ten: and this it is
that puzzles me and many wiser men.

I think that maybe the ten men were never in a room all at the same time or
maybe there was a set of siamese twins