You are in the desert and you have 3 buckets of water containing a,b,c liters respectively (a,b,c - positive integers).
You need an empty bucket for an unspecified purpose. Being in the desert you need the water and cannot just pour it away.
You have to pour the contents of one bucket into another one. But in any pouring, you must double the contents of the bucket which receives the water.
For example the sequence of bucket contents could be:
3 2 1
1 4 1
0 4 2
Now show that no matter what a,b,c are, you can always manage to empty a bucket under this constraint.
You may assume:
a>b>c
&
(capacity of each bucket)>(a+b)
(In reply to
Still stumped by Steve Herman)
You seem to have gotten about as far as I have. But I noticed one more thing in the two bucket scenario. From what I have seen if the two buckets are coprime along with having opposite parity then the cycle contains all possible distributions. For example (1,10) -> (2,9) -> (4,7) -> (8,3) -> (5,6) -> (10,1) etc.
If this is indeed true for all such pairs then the puzzle is solved by making the largest and one other of the three buckets have opposite parity and be coprime. Then apply the two bucket cycle to the pair until the third bucket's volume appears in the cycle. At that point two buckets are equal and the final transfer can be done.
re: Still stumped
