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Five men and a monkey were shipwrecked on a island, and they spent the first day gathering coconuts.They piled them up together and then went to sleep for the night.

But when they were all asleep one man woke up, and he thought there might be a row about dividing the coconuts in the morning, so he decided to take his share. So he divided the coconuts into file piles. He had one coconut left over, and he gave that to the monkey, and he hid his pile and put the rest all back together.

By and by the next man woke up and did the same thing. And he had one left over and he gave it to the monkey. And all five men did the same thing, one after the other; each one taking a fifth of the coconuts in the pile when he woke up, and each one having one left over for the monkey.

And in the morning they divided what coconuts were left and they came out in five equal shares. Of course each one knew that there were coconuts missing but noone spoke up.

How many coconuts were there in the beginning?

2006-08-04 17:13:29 · 5 answers · asked by Will 4 in Science & Mathematics Mathematics

LOL. If I told you that the answer would be trivial. You can solve it without that knowledge.

Hint: It is an integer and has 5 or less digits.

2006-08-04 17:29:45 · update #1

Yes David. I ran out of room to describe the problem. Each person was only dividing the main pile into five equal piles (for the people) and then giving the one leftover coconut to the monkey.

2006-08-04 17:44:30 · update #2

Well done again pascal. Someone knows how to google the problem to get the answer.

Another solution though it is kind of odd is: -4

Each man divides the pile into 5 equal piles of -1 coconut each and then tosses the extra "positive" coconut to the monkey. He then takes out his "-1" coconut and returns the remaining -4 coconuts to the pile. Each man does and winds up with -2 coconuts (-1 in middle of night and -1 in morning) while the monkey gets a +6 coconuts.

2006-08-04 17:52:22 · update #3

5 answers

3121 coconuts.

Each man gives one coconut to the monkey and then leaves 4/5 of what remains. Therefore, if what remains is an integer number of coconuts, it must be divisible by four. Further, if what remains after N men have woken up is x, then what remained after N-1 men have woken up is 5x/4 + 1, and this must also be divisible by four. Thus, the final number of coconuts must be itself divisible by four and remain so throughout 4 iterations of f(x)=5x/4+1. It must also be divisible by 5, which means the final number must be divisible by 20. If such a number exists, 5 iterations of such a number through f(x)=5x/4+1 gives the original number of coconuts. A brute-force search reveals that the smallest number fulfilling these criteria is 1020, which would remain from an inital stock of 3121 coconuts.

Edit: I am shocked at the implication that I used google to solve this problem. That was 100% original reasoning and tedious calculation.

2006-08-04 17:44:37 · answer #1 · answered by Pascal 7 · 0 1

-5 coconuts.
The first man woke up and divided the -5 coconuts into 5 piles of -1 each. He took one pile and tossed a positive 1 coconut to the monkey, leaving -5 coconuts again.
Each man in turn did the same, leaving -5 coconuts each time.
When they woke up, the 5 men split the -5 coconuts.

Note: It would be -4 if they gave the monkey one last coconut after they woke up. but you didnt say that. You cant divide -4 evenly at the end unless you give the monkey one.

2006-08-04 18:01:01 · answer #2 · answered by Scott R 6 · 0 0

I dunno...
x
((((((((((x*5)+1)*5)+1)*5)+1)*5)+1)*5)+1)
x is a multiple of 5... lets say its 16,406 using the smallest value of left over coconuts...

*edit* x was referring to the coconuts left after they woke up and each have taken a share. I did not understand the wording fully, but I assumed that each person only divided the coconuts up for the other humans, and just left the final one for the monkey as a present or something :-)

*edit* So... did I get it right?

*edit* No, I didn't... I just tried it and got fractions of coconuts ^_^

*edit* I wonder what I did wrong... you have x coconuts at the end, where 5 is divisible by 5... (x*5)+1 should be the number of coconuts there before the last person took his share... Oh... Ohhh....... So it should be (((x-1)*1.25)+1) and since x has to be a multiple of 5...ugh... now it has to be a multiple of 5 by a multiple of 4 minus 3... *sighs* I'm missing something, I'll get back to you later.

*edit* Ahh... that is a good idea Pascal, realizing that each must be divisible by 4.

2006-08-04 17:40:16 · answer #3 · answered by Anonymous · 0 0

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2016-12-11 07:02:45 · answer #4 · answered by Anonymous · 0 0

erm...30?I don't noe about this?

2006-08-04 17:33:08 · answer #5 · answered by Anonymous · 0 0

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