You have a 12 litre jug, an 8 litre jug, and a 5 litre jug. None of the jugs have any markings on them. The 12 litre jug is full, and the other two are empty.

How can you divide the 12 litres of water in such a way that two of the jugs (the 12 litre and the 8 litre jugs) have exactly 6 litres of water in them, and the third (the 5 litre jug) is empty?

The solution to this works pretty much round robin from 12 to 5 to 8. Once 8 is full, empty 8 back to 12, then whatever is left in 5 goes into 8. Then restart the cycle from 12 to 5 to 8 until 8 is full, empty 8 to 12, then remaining contents of 5 to 8, and start again.

Start with contents 12-0-0 (jugs in capacity order of 12-8-5)
1. From 12 to 5 = 7-0-5
2. From 5 to 8 = 7-5-0
3. From 12 to 5 = 2-5-5
4. From 5 to 8 = 2-8-2
5. 8 is full, so from 8 to 12 = 10-0-2
6. Empty remaining from 5 to 8 = 10-2-0
7. Restart cycle like at #1, i.e. 12 to 5 = 5-2-5
8. 5 to 8 = 5-7-0
9. 12 to 5 = 0-7-5
10. 5 to 8 = 0-8-4
11. 8 is full, so from 8 to 12 = 8-0-4
12. Empty remaining from 5 to 8 = 8-4-0
13. Restart Cycle like at #1 & #7, so 12 to 5 = 3-4-5
14. 5 to 8 = 3-8-1
15. 8 is full, so empty 8 into 12 (almost done)= 11-0-1
16. Empty remaining 5 to 8 = 11-1-0
17. Restart 12 to 5 = 6-1-5
18. Empty 5 into 8 = 6-6-0, and we are DONE!

Posted by Lawrence on 2003-08-23 05:50:45