The following special catch puzzle appeared in the issue of The Weekly Dispatch for All Fools' Day, 1900. It caused considerable amusement; for out of a very large body of competitors, many quite expert, not a single person solved it, though it ran for nearly a month.

" A race between a man and a woman that I happened to witness one All Fools' Day has fixed itself indelibly on my memory. It happened at a country-house, where the gardener and the cook decided to run a race to a point 100 feet straight away and return. I found that the gardener ran 3 feet at every bound and the cook only 2 feet, but then she made three bounds to his two. Now, what was the result of the race?"

A fortnight after publication the editor added the following note: "It has been suggested that perhaps there is a catch in the 'return,' but there is not. The race is to a point 100 feet away and home again—that is, a distance of 200 feet. One correspondent asks whether they take exactly the same time in turning, to which I reply that they do. Another seems to suspect that it is really a conundrum, and that the answer is that 'the result of the race was a (matrimonial) tie.' But I had no such intention. The puzzle is an arithmetical one, as it purports to be."

While it would seem that they will stay on pace with each other throughout the race, the fact that they have to run to a point 100 feet away and turn around changes things.  100 is not evenly divisible by three - the gardener will actually run to a point 102 feet away (on their 34th bound).  The cook, not having this problem, will have by that point already turned around and started heading back and will therefore win the race by a slight margin.

If that is not the solution, then is it possible that the gardener is the female?

 

Posted by tomarken on 2006-03-01 13:36:35