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You have three buckets, two big buckets holding 8 litres of water each and one small empty bucket that can hold 3 litres of water. How will you split the 16 litres of water to four people evenly? Each person has a container but once water is distributed to someone it cannot be taken back.

In this puzzle, we need to allocate 4 litres to each person. So I considered the initial state as below

8 8 0 [0, 0, 0, 0]

The values in the bracket are the water given to those 4 people. I tried the below steps

5 8 3 [0, 0, 0, 0]
5 8 0 [3, 0, 0, 0]
5 5 3 [3, 0, 0, 0]
5 5 0 [3, 3, 0, 0]
2 8 0 [3, 3, 0, 0]
0 8 2 [3, 3, 0, 0]
0 7 3 [3, 3, 0, 0]
0 4 3 [3, 3, 3, 0]
0 1 3 [3, 3, 3, 3]
0 0 3 [4, 3, 3, 3]

But I couldnt take it further by giving those 1 litre for rest of the 3 persons since there is a constraint that water given to person cant be retrieved back.
Any help/hint towards final solution is greatly appreciated.

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To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. –  Zev Chonoles Mar 8 '13 at 10:32
    
@ZevChonoles Sure. Let me update the question now. –  sundar Mar 8 '13 at 10:36
    
@sundar: Is the water in the big buckets to begin with, or is it in another container? –  Dennis Gulko Mar 8 '13 at 10:39
    
@Dennis: The question is rather clear in that respect: The big buckets are "holding" water and the small one is empty and "can hold" water. –  joriki Mar 8 '13 at 10:43
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1 Answer 1

up vote 7 down vote accepted
8 8 0 [0 0 0 0]
8 5 3 [0 0 0 0]
8 5 0 [3 0 0 0]
8 2 3 [3 0 0 0]
8 0 3 [3 2 0 0]
8 3 0 [3 2 0 0]
5 3 3 [3 2 0 0]
5 6 0 [3 2 0 0]
2 6 3 [3 2 0 0]
2 8 1 [3 2 0 0]
2 8 0 [3 2 1 0]
0 8 2 [3 2 1 0]
0 7 3 [3 2 1 0]
3 7 0 [3 2 1 0]
3 4 3 [3 2 1 0]
6 4 0 [3 2 1 0]
6 1 3 [3 2 1 0]
6 0 3 [3 2 1 1]
8 0 1 [3 2 1 1]
8 0 0 [4 2 1 1]
5 0 3 [4 2 1 1]
5 3 0 [4 2 1 1]
2 3 3 [4 2 1 1]
0 3 3 [4 4 1 1]
0 0 3 [4 4 4 1]
0 0 0 [4 4 4 4]
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+1 Nice! Any particular approach to do this, better than trial and error? –  polkjh Mar 8 '13 at 11:13
    
@polkjh: The only insights I remember having are a) I was always trying to do it with $3+1$ and I needed to include some $2$s, and b) a $1$ or $2$ needed to be dispensed whenever possible, preferably a $1$, to produce as many $1$s and $2$s as possible early on, since that's no longer possible once you only have $6$ litres left. –  joriki Mar 8 '13 at 11:21
    
Is the 6th line redundant? –  John Bentin Mar 8 '13 at 13:38
    
@John: You mean you could go directly from the $5$th to the $7$th? I don't think so. The first two numbers stand for the big buckets (as in the OP's notation); you can't just pour $3$ litres from one big bucket into the other without using the small bucket to measure them. –  joriki Mar 8 '13 at 13:52
    
Got it! Thank you! –  John Bentin Mar 10 '13 at 9:31
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