# Three Jugs - Equal Amounts

^{n}pints of water. Each jug not empty and contains an integer amount of pints. We may pour into any jug as much water as it already contains, from any other jug with a greater amount of water. Prove that, with one exception, it is possible after a series of such pourings to have the same amount of water in all three jugs.

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Copyright © 1996-2017 Alexander Bogomolny

^{n}pints of water. Each jug not empty and contains an integer amount of pints. We may pour into any jug as much water as it already contains, from any other jug with a greater amount of water. Prove that, with one exception, it is possible after a series of such pourings to have the same amount of water in all three jugs.

### 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 3·2^{k+1} pints of water distributed among several jugs. We shall first show that, excluding one exceptional distribution of water, it is always possible to get even amounts of pints in all three jugs.

If there are jugs with an odd amount of pints, there must be 2 of them. There are two possibilities. Let the odd amounts be a (in jugs A_{1} and A_{2}) and the remaining even one b (in jug B).

If a < b, pour from B to, say, A_{1} to obtain distribution of even numbers

If a > b, move from the distribution (a, a, b) to (a, a - b, 2b). A repetition of the previous configuration is not possible. For then we would have ^{k+1}. On the next pouring, move to

Once all the jugs contain even amounts of water, we may imagine the pint unit increased by a factor of two, making the total 3·2^{k} of "big" pints and reducing the case to the inductive assumption.

The only exceptional case where the argument does not work is the configuration (a, a, 2a), for which ^{k+1}.

Now note that we came across the exceptional case

To sum up, the problem is solvable for the total of 3·2^{n} pints of water, unless the initial distribution is ^{n-2}, 3·2^{-2}, 3·2^{n-1}),

As an example, let's have three jugs A, B, C with amounts of water

A | B | C | |
---|---|---|---|

7 | 7 | 10 | |

7 | 14 | 3 | C→B |

7 | 11 | 6 | B→C |

1 | 11 | 12 | A→C |

1 | 22 | 1 | C→B |

2 | 21 | 1 | B→A |

2 | 20 | 2 | B→C |

4 | 18 | 2 | B→A |

4 | 16 | 4 | B→C |

8 | 12 | 4 | B→A |

8 | 8 | 8 | B→C |

What could be a converse theorem?

- 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 © 1996-2017 Alexander Bogomolny

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