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Problems of this type have been around for a long time,
but only recently did I realize that they are Diophantine equation
problems in disguise.
If we add 7's and subtract 11's, eventually we'll get to 6, or whatever other number is desired.
We just need to solve 7x - 11y = 6
I have another page which does that, but here I'll work this one out.
7 (x - y) - 4y = 6.
Let w = (x-y)
7w - 4y = 6
3w + 4 (w - y) = 6
Let v = w - y
3w + 4v = 6
We can solve this by inspection: v = 0, w = 2.
Then work backwards:
0 = 2 - y
y = 2
x - y = 2, x = 4
7 * 4 - 11 * 2 = 28 - 22 = 6.
If we fill the 7 liter jug 4 times,
and empty the 11 liter jug twice,
we'll have our 6 liters.
We proceed as follows:
| Action | Count | 7L jug | 11L jug |
|---|---|---|---|
| Start | 0 | 0 | |
| Fill 7 | (1) | 7 | 0 |
| Pour into 11 | 0 | 7 | |
| Fill 7 | (2) | 7 | 7 |
| Pour into 11 | 3 | 11 | |
| Empty 11 | [-1] | 3 | 0 |
| Pour into 11 | 0 | 3 | |
| Fill 7 | (3) | 7 | 3 |
| Pour into 11 | 0 | 10 | |
| Fill 7 | (4) | 7 | 10 |
| Pour into 11 | 6 | 11 | |
| At this point we have our 6, and don't need to carry out the last step, but it fits the equation better if we do. | |||
| Empty 11 | [-2] | 6 | 0 |
