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# Kit kat sharing problem watch

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1. You have a four piece kit kat (not sure if you get them anymore, but they were great btw) and have five friends. You can only split the kit kat into equal halfs- for example, 1/2 then 1/4 then 1/8 nothing else. You can give your friends as many seperate pieces as possible, but all of them have to get an equal amount.

Is it possible to give an equal amount of all of them, and if so, how?

My friend solved this using proof by induction, I would like to see if he was correct
2. btw, all of the kit kat has to be used
3. You do still get them I think.

I don't understand the problem - are you saying you can only split the 4 into powers of 1/2, so 1/64 would be acceptable? Or that you can split each individual piece into 1/2, 1/4 or 1/8 ONLY? Or something else all together?
4. you can only split any piece you have into two seperate halfs. so powers of 1/2 is a better way of saying it yeah
5. Tell them to **** off and buy their own chocolate bars.
6. You could get infinitely close to giving everyone one fifth, but never exactly right... are you asking for the sum of fractions that each person should get?
7. No, can't be done.
8. Impossible. If I can't do it with 8 pieces, there is no point in carrying on.
9. (Original post by tomthecool)
You could get infinitely close to giving everyone one fifth, but never exactly right... are you asking for the sum of fractions that each person should get?

If can only get close then its impossible so you've answered the question - nice one.

But can you explain why only infinitely close?
10. I don't think it's possible. You're effectively saying that there's some (finite, presumably) n and x such that , which won't ever happen.
11. And no powers of 1/2 have a denominator divisible by 5
12. ah yes.

2^n = 5x
now if we assume that statment is true, can we rewrite x as 2y because x has to be even to produce an even number:
2^n = 5 x 2y

2^(n-1) = 5y
Therefore if its true for any power of 2 its true for all of them
clearly this isn't true; therefore 2^n can not = 5x
13. Can I have a whole finger please?
14. (Original post by hai2410)
If can only get close then its impossible so you've answered the question - nice one.

But can you explain why only infinitely close?
Short and simple answer: Because 2 and 5 are coprime.

Or to put it mathematically:
Each person gets x pieces of the bar, after it has been divided in two n times, such that the sum of all their pieces is 1.

But since x and n are integers, 5x must be a multiple of 5, and 2^n cannot be a multiple of 5. Therefore the two values can never (quite) be equal.
15. or you could split lengthways into five pieces, instead of being stupidly pedantic
16. Try showing you can't take a fifth off one chocolate bar. Maybe think of the factors of the number of cuts made with halving the number of cuts each time.
17. Kit Kat Chunky's don't have this problem.
18. (Original post by hai2410)
You have a four piece kit kat (not sure if you get them anymore, but they were great btw) and have five friends. You can only split the kit kat into equal halfs- for example, 1/2 then 1/4 then 1/8 nothing else. You can give your friends as many seperate pieces as possible, but all of them have to get an equal amount.

Is it possible to give an equal amount of all of them, and if so, how?

My friend solved this using proof by induction, I would like to see if he was correct
Yeah, you can. After you break, check if giving it to the person would exceed 1/5th. If it doesn't, give, then break further. If it does, don't give, just break. Performing this algorithm will give him a increasing sequence of parts which converges to 1/5th.
19. (Original post by The Bachelor)
Yeah, you can. After you break, check if giving it to the person would exceed 1/5th. If it doesn't, give, then break further. If it does, don't give, just break. Performing this algorithm will give him a increasing sequence of parts which converges to 1/5th.
Converge after an infinite number of splits though? ?
20. (Original post by hai2410)
Converge after an infinite number of splits though? ?
Yes. This is possible if, for example, the nth split took seconds. It's kind of vague what the capabilities of the splitter are.

By the way, even with all of your restrictions, there is a trivial solution that satisfies all given conditions. You give each of your friends 0, and eat the entire bar.

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Updated: November 17, 2008
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