You have a 3 and a 5 litre water container, each container has no markings except for that which gives you it's total volume. You also have a running tap. You must use the containers and the tap in such away as to exactly measure out 4 litres of water. How is this done?

There are two ways to solve this, maybe the question could be modified to say the 5 litre can doesn't fit under the tap...

**Number 1**

Fill the 5 litre can from the tap

Empty the 5 litre can into the 3 litre can - leaving 2 litres in the 5 litre can.

Pour away the contents of the 3 litre can.

Fill the 3 litre can with the 2 litres from the 5 litre can - leaving 2 litres in the 3 litre can.

Fill the 5 litre can from the tap.

Fill the remaining 1 litre space in the 3 litre can from the 5 litre can.

Leaving 4 litres in the 5 litre can.

**Number 2**

Fill the 3 litre can from the tap.

Empty the contents of the 3 litre can into the 5 litre can.

Fill the 3 litre can from the tap.

Empty the contents of the 3 litre can into the 5 litre can. - Leaving the 5 litre can full and 1 litre in the 3 litre can.

Pour away the contents of the 5 litre can

Pour the 1 litre from the 3 litre can into the 5 litre can.

Fill the 3 litre can from the tap.

Empty the contents of the 3 litre can into the 5 litre can.

Leaving 4 litres in the 5 litre can.

## Generalised Form

We have two solutions above which both give the same answer of 3 litres. It turns out that if you make it algebraic the answer does not have the same form and it's sort of coincidence. Lets call the smaller container A and the lager B and work it through.**Number 1**: Empty the 5 litre can into the 3 litre can - leaving 2 litres in the 5 litre can. ie B - A in B

Fill the 3 litre can with the 2 litres from the 5 litre can - leaving 2 litres in the 3 litre can. As in space in A of A - (B - A) = 2A - B

Fill the remaining 1 litre space in the 3 litre can from the 5 litre can. Leaving

__4 litres__in the 5 litre can. So the amount in B is B - (2A - B) or 2B - 2A

So the generalised form of solution 1 is 2B - 2A in B.

**Number 2**: Fill the 3 litre can from the tap. Empty the contents of the 3 litre can into the 5 litre can. Which gives us B - A space in B.

Fill the 3 litre can from the tap. Empty the contents of the 3 litre can into the 5 litre can. - Leaving the 5 litre can full and 1 litre in the 3 litre can. ie. removing (B - A) from A leaves A - (B - A) or 2A - B in A

Pour away the contents of the 5 litre can. Pour the 1 litre from the 3 litre can into the 5 litre can. ie 2A - B in B

Fill the 3 litre can from the tap. Empty the contents of the 3 litre can into the 5 litre can. Leaving

__4 litres__in the 5 litre can. ie A + (2A - B) or 3A - B in B.

So the generalised form of solution 2 is 3A - B in B.

There are some restrictions, such as in solution 2; 2A < B (2*A must be less than B,) but other wise we can make up other puzzles like 5 Litres and 9 Litres to get 6 Litres. or 8 Litres. All exactly the same form.

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