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Topic: Diophantus and the coconuts (Read 1196 times) |
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Diophantus and the coconuts
« on: Feb 26th, 2004, 6:37pm » |
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I just picked this one up from an old Omni book.
I solved it for one value, but have not yet looked at the answer page (for me this is an incredible show of restraint).
Five sailors are shipwrecked on an island. On the island they find one monkey, and the only supply of food is coconuts.
They collect as many coconuts as possible and put them in a big pile together. They are tired so they all go to sleep.
However, sailor Angus wakes up during the night and is suspicious of the other sailors taking his share of the coconuts. So, Angus divides the coconuts up into five piles. But, there is one coconut left over, which he gives to the monkey. Then he takes his share and hides it. Then he puts all the remaining coconuts back into one pile and goes back to sleep.
An hour later, the second sailor Bangus wakes up and does the same thing. Divides the coconuts into five piles, with one coconut left over that he gives to the monkey. Then he hides his share of the coconuts and goes to sleep.
Cangus, Dangus and Eugene all do the same thing, and always there is one coconut remaining.
They all wake up in the morning and notice that the pile is smaller, but all being guilty they say nothing. They divide the pile five ways and again find one coconut left over which they give to the monkey.
How many coconuts did they start with?
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Cathos
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Re: Diophantus and the coconuts
« Reply #1 on: Feb 26th, 2004, 7:31pm » |
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I remember seeing this riddle eslewhere. I don't remember the exact answer but it's much more than one would think.
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Sum Arbor
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Re: Diophantus and the coconuts
« Reply #2 on: Feb 26th, 2004, 7:37pm » |
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I think it is too. I have to take it back. I tried to answer it but my first try was based on bad math. I am trying again. 
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John_Gaughan
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Re: Diophantus and the coconuts
« Reply #3 on: Feb 26th, 2004, 9:13pm » |
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:: For n < 100,000 I get:
15621
31246
46871
62496
78121
93746 ::
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x = (0x2B | ~0x2B)
x == the_question
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Speaker
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Re: Diophantus and the coconuts
« Reply #4 on: Feb 26th, 2004, 9:42pm » |
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Okay, now I have an answer. I hope it is correct. I will look at the book only if there is some agrument.
31,246 coconuts, of which six went to the monkey, these guys worked pretty hard, but were also pretty stingy.
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ThudnBlunder
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Re: Diophantus and the coconuts
« Reply #5 on: Feb 26th, 2004, 9:47pm » |
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Let n = the original number of coconuts.
Let m = how many coconuts each sailor receives in the morning.
:Then we end up needing to find the smallest positive integers which satisfy 1024n - 15625m = 11529
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| « Last Edit: Feb 27th, 2004, 2:55am by ThudnBlunder » |
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Barukh
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Re: Diophantus and the coconuts
« Reply #6 on: Feb 27th, 2004, 1:29am » |
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Here's a nice way to approach the problem.
[smiley=blacksquare.gif]
Note that if N is a solution, then N + 56 is also a solution.
Next note that, formally, -4 (minus four) coconuts satisfy all the conditions. Thus, we get the smallest number N = -4 + 56 = 15621. Repeating, we get all the numbers in John's list.
Also, it is interesting to represent the answer in base 5: 15621 = 4444415. It makes the "division" process more transparent.
[smiley=blacksquare.gif]
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| « Last Edit: Feb 27th, 2004, 1:30am by Barukh » |
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NickH
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Re: Diophantus and the coconuts
« Reply #7 on: Feb 27th, 2004, 1:25pm » |
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Here's another approach...
Let the original pile have n coconuts. Let a be the number of coconuts in each of the five piles made by the first man, b the number of coconuts in each of the five piles made by the second man, and so on. Then
n = 5a + 1 => n + 4 = 5(a + 1)
4a = 5b + 1 => 4(a + 1) = 5(b + 1)
4b = 5c + 1 => 4(b + 1) = 5(c + 1)
4c = 5d + 1 => 4(c + 1) = 5(d + 1)
4d = 5e + 1 => 4(d + 1) = 5(e + 1)
4e = 5f + 1 => 4(e + 1) = 5(f + 1)
Hence n = 5*(5/4)5(f + 1) - 4 = (56/45)(f + 1) - 4.
Since 5 and 4 are coprime, 56/45 = 15625/1024 is a fraction in its lowest terms, and so the only integer solutions are where f + 1 is a multiple of 45.
So the general solution is n = 15625r - 4, where r is a positive integer, giving a smallest solution of 15621 coconuts in the original pile.
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Re: Diophantus and the coconuts
« Reply #8 on: Feb 29th, 2004, 4:01pm » |
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Well, you guys are correct, of course you do not need me to tell you that. But I do have the book. My answer, which was correct, but not the lowest possible answer was 31,246 coconuts in the original pile.
To get my answer I used an excel worksheet and started at 10,001 and made a list of integers 9,996, 9991, 9986 etc. Figuring that the first number had to be divisible by 5 plus 1. Well, I ran through all those calculations and then scrolled through the answer looking for the answer that held no digits after the decimal point. I didn't find any so I went up a little higher. So when I scrolled down the first possible correct answer I found was higher than the correct answer.
I made the same mistake with the sell the beer and drink one at the end of the day riddle. Oh well.
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| « Last Edit: Feb 29th, 2004, 4:04pm by Speaker » |
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