Three Jugs - Equal Amounts

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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 3·2k+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 A1 and A2) and the remaining even one b (in jug B).
If a < b, pour from B to, say, A1 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
Once all the jugs contain even amounts of water, we may imagine the pint unit increased by a factor of two, making the total 3·2k 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
Now note that we came across the exceptional case
To sum up, the problem is solvable for the total of 3·2n pints of water, unless the initial distribution is
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

|Contact| |Front page| |Contents| |Arithmetic| |Math induction|
Copyright © 1996-2018 Alexander Bogomolny
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