# Many Jugs to One I

This is a problem from a Bulgarian 1989 winter competition. The proof below, by Miroslav Petkov, was awarded a special prize.

### References

- S. Savchev, T. Andreescu,
*Mathematical Miniatures*, MAA, 2003

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Copyright © 1996-2018 Alexander Bogomolny

### Proof

Assume on the contrary that n = 2^{k}m, where m > 1 is odd. Choose such an initial distribution of water as to have every jug contain a multiple of 2^{k}. It is always possible by, say, having 2^{k} in one jug and ^{k}(m - 1)^{k} pints. Before the very last pouring, there will be two jugs with equal amounts of water, say 2^{k}s. After the last pouring, one of the jugs will contain ^{k}s + 2^{k}s = 2^{k+1}s

^{k}m = n = 2

^{k+1}s,

making m even. A contradiction.

(The problem admits a converse.)

- 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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