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."

I didn't see this in the solutions presented so far, so I'll try posting it.

The result is a tie.

The reason for this is that this is an "arithmetical" puzzle. I assume this to mean that all this complication about non-partial bounds and turning is meant to be irrelevant. It's like those word problems that are given in grade school. Real life details/physics don't count.

The other thing is semantics, which is harder to prove since the rules of proper/natural grammar can be quite subjective. For simplicity's sake, I assume the "but then" makes it clear that the woman is the cook.

Based on this, it can be calculated that it would take the man 66 and two-third bounds to complete the 200 feet (200ft divided by 3ft/bound).

It would take the woman exactly 100 bounds to complete 200 feet (200ft divided by 2ft/bound).

Since the man only goes 2 bounds for the woman's every 3 bounds, the number of bounds the woman makes for the man's bounds can be calculated as: "66 2/3 multiplied by 3/2" (i.e, 66 2/3 man's bounds multiplied by 3 woman's bounds per 2 man's bounds).

The calculation turns out to be 100 bounds for the woman, the exact same number of bounds it takes her to travel the same distance. Thus, it's a tie.

Posted by Monika on 2006-03-03 11:33:43