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Topic: st. edmundsbury's riddle (Read 1764 times) |
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Moti
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st. edmundsbury's riddle
« on: Nov 24th, 2002, 1:45pm » |
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Long ago in st. edmundsbury monastery, as father peter told on one occasion, they were flooded with so many mice that the head of the monastery ordered to gather all the cats from the area to get rid of the rodents.
records were kept and on the end of the year they found that
exactly 1,111,111 mice were killed.
if every cat killed an equal number of mice, try and say how many cats were envolved.
(could be more than one answer)
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| « Last Edit: Nov 25th, 2002, 8:43am by Moti » |
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Garzahd
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Re: st. edmundsbury's riddle
« Reply #1 on: Nov 24th, 2002, 5:52pm » |
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Must the cats be equally ravenous?
If so, then you're just asking for the prime factorization of that number, which is 1 (hungry kitty!), 239, 4649, 1111111 (presenting a new sort of infestation problem).
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Moti
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Re: st. edmundsbury's riddle
« Reply #3 on: Nov 25th, 2002, 12:57am » |
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I'm very sorry guys.
i was very tired when writing this.
of course every cat killed an equal number of mice.
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william wu
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Re: st. edmundsbury's riddle
« Reply #4 on: Nov 25th, 2002, 8:32am » |
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All cats eat the same number of mice. Thus we want to evenly partition the group of mice such that every cat is allocated the same integer number of mice (assuming that no mouse was jointly fragged by more than one cat). So if we can figure out the size of a partition, we can divide 1111111 by that number to get the number of cats. Put more simply, we just need to know what integers divide evenly into 1111111. We can factorize the number 1111111 to get possible partition sizes:
>> factor(1111111)
ans =
239 4649
1111111 happense to have only these two large prime factors, which makes our list of possible answers short.
If every cat eats 239 mice, then there were 4649 cats involved in the purge.
If every cat eats 4649 mice, then there were 239 cats involved in the purge.
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Moti
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Re: st. edmundsbury's riddle
« Reply #5 on: Nov 25th, 2002, 8:39am » |
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God bless you, William wu. 
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icon
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Re: st. edmundsbury's riddle
« Reply #6 on: Nov 25th, 2002, 8:50am » |
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william,
did you do this by hand or there is some program which bruteforces them?
ciao
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Moti
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Re: st. edmundsbury's riddle
« Reply #7 on: Nov 25th, 2002, 8:55am » |
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although its been a long time since i did any programing.
i suppose you can just write a program that divides
a given number by a growing integer.
and gives you only the numbers that produced whole numbers
by division.
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towr
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Re: st. edmundsbury's riddle
« Reply #8 on: Nov 25th, 2002, 9:09am » |
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There are programs that can factor in a smart way..
Even the trial version of Derive 5 can do it..
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Wikipedia, Google, Mathworld, Integer sequence DB
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Jeremiah Smith
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Re: st. edmundsbury's riddle
« Reply #9 on: Nov 25th, 2002, 10:55pm » |
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Well, you could just have 1,111,111 cats who each kill one, or, alternatively, have a single überCat who killed all 1,111,111 (Well, the puzzle description does specify cats plural...darn...I was looking forward to seeing the überCat.  )
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william wu
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Re: st. edmundsbury's riddle
« Reply #10 on: Dec 8th, 2002, 3:51pm » |
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icon: I used the canned factor() function in MATLAB, which uses the sieve method.
jeremiah: Oops I forgot those cases. Yea, nice catch.
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jon_G
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Re: st. edmundsbury's riddle
« Reply #11 on: Dec 8th, 2002, 6:28pm » |
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At first I thought this was a trick question but after a little thinking I knew it wasn't. My first thought was 1,111,111cats. I love the ubercat. 
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