# The Seven problem

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165330.  Thu Apr 12, 2007 10:47 am

Can't think where to put this:

A well-known medieval riddle:

 Quote: As I was going to St. Ives, I met a man with seven wives. Every wife had seven sacks; Every sack had seven cats; Every sack had seven kittens. Kittens, cats, sacks and wives: How many were going to St. Ives?

The trick is that the answer is '1', as it's only the man that's actually going there.

There is a striking resemblance of this riddle to Problem 79 (the 'sevens problem') of the Rhind Mathematical Papyrus, copied into Hieratic by the Egyptian scribe Ahmes from a 12th Dynasty (roughly 2000 - 1800 BC) original, now lost.

 Quote: Houses: 7 Cats: 49 Mice: 343 Spelt: 2401 (a type of wheat) Hekat: 16807 (a volume unit for grain) The total: (19607)

Leonardo of Pisa, also known as Fibonacci, copied this puzzle into his Liber Abaci in 1202, and it may also be the source of the Hebrew parable:
 Quote: This is Yavah who vanquished Death which killed The butcher which slew The ox which drank The water which quenched The fire which burnt The stick which beat The dog which worried The cat which killed The kid which my father bought for two zuzim.

and the fable:
 Quote: For want of a nail the shoe was lost. For want of a shoe the horse was lost. For want of a horse the rider was lost. For want of a rider the battle was lost. For want of a battle the kingdom was lost. And all for the want of a horseshoe nail.

and the nursery rhyme 'The House That Jack Built', popularised by Randolph Caldecott.

The final conditions are all 'contingent properties' because they all rely very heavily on the sensitivity of the first condition.

The Dictionary of Curious and Interesting Numbers
Gillings, Mathematics of the Pharaohs, MIT, 1972
Brewer's Phrase and Fable

 165341.  Thu Apr 12, 2007 10:58 am Enigmas?

 165343.  Thu Apr 12, 2007 11:00 am Chris - I guess I'm being dim here but what is the question in Problem 79? I can't see one.

 165362.  Thu Apr 12, 2007 11:28 am Sorry - I didn't make that clear. Problem 79 is a 'grouping problem' - i.e. how do you go about adding up all those numbers efficiently? Aside from the 'St Ives' version being a trick (the answer is 1), the papyrus goes into some detail explaining that you don't need to add up all the numbers - there's a 'short cut' for working out the sum of a geometric series, if you group the numbers together cleverly. It's like the old chestnut: "What's the sum of all the numbers from 1 to 100?" This is, strictly speaking, an arithmetic series, but there's a similar kind of shortcut. You could doltishly add 1, 2, 3, 4... for a few hours, or you could cunningly group the numbers thusly: (1 + 99) + (2 + 98) + (3 + 97)... until you're left with 50 by itself. The answer is then (49 * 100) + 50 = 4950. http://mathworld.wolfram.com/StIvesProblem.html

 165367.  Thu Apr 12, 2007 11:39 am Oh, OK. I have to say, it looks to me to be completely different from the St Ives thing, in that it isn't a riddle at all.

 165395.  Thu Apr 12, 2007 1:15 pm It's not phrased in quite the same way, but it's an identical mathematical problem. Agreed, most riddles aren't mathematical (unless you count strict logic), but it's definitely the same. Probably not question material, though.

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