Many Jugs to One II
(The is a converse of another problem.)
References
- S. Savchev, T. Andreescu, Mathematical Miniatures, MAA, 2003
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Copyright © 1996-2018 Alexander Bogomolny
Proof
The proof is by induction on n. The statement is clearly true for n = 1. Assume it holds for n = k and let there be 2k+1 pints of water distributed among several jugs. The number of the jugs that contain an odd amount of pints must be even. Arrange them in pairs and pour from one jug in a pair into another. This will make the number of pints in each of the given jugs even. From then on every jug will contain an even number of pints.
The problem is then equivalent to the one where the jugs contain half of what they contained on the previous stage to the total of 2k. By the inductive assumption, there is a sequence of pourings that leaves a single jug with all the water and other jugs empty.
Returning to the original configuration, we may simply repeat the same sequence of pourings but with twice the amount.
- The Three Jugs Problem. Introduction and a story
- 3 Glasses Puzzle
- Water puzzle, experimental math
- Three Glass Puzzle (Graph Theoretical Approach)
- The puzzle in barycentric coordinates
- Two Pails Puzzle
- Plain Gadgets
- 3 Jugs Problem - A Water Doubling Variant
- Many Jugs to One I
- Many Jugs to One II
- Three Jugs - Equal Amounts
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Copyright © 1996-2018 Alexander Bogomolny
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