We saw in class how to solve the Die Hard 3 water problem: given a 5 gallon jug and a 3 gallon jug. The solution ultimately involved filling the 5 gallon jug twice and emptying the 3 gallon jug 2 twice, giving 4 = 2*5 - 2*3 gallons. In more detail, we filled the 5 gallon jug, used that to fill the 3 gallon jug which we dumped out giving us 5 - 3 = 2 gallons in the 5 gallon jug. We poured those 2 gallons into the now empty 3 gallon jug so we could then fill the 5 gallon jug a second time. Again we used the water in the 5 gallon jug to fill the 3 gallon jug (which you remember already has 2 gallons in it). This gave us 4 gallons in the 5 gallon jug, and we dumped out the water in the 3 gallon jug leaving us with 4 gallons altogether. But even though there was a certain amount of moving the water back and forth in the two jugs, ultimately we filled the 5 gallon jug twice and emptied the 3 gallon jug twice, giving us 4 = 2*5 - 2*3 gallons in the end.

This solution first involved writing 4 a multiple of 5 minus a multiple of 3. We also saw in class how to obtain other exact quantities of water:

2 = 5 - 3

3 = 0*5 + 1*3

4 = 2*5 - 2*3

5 = 1*5 + 0*3

6 = 0*5 + 2*3

8 = 1*5 + 1*3

We didn't in class decide whether we can get 1 or 7, but we did notice that we can't get more than 8 gallons, because we have only the two jugs and they add up to a total volume of 8 gallons.

In each of these cases, the arithmetical expression can be converted into an actual real life way of getting an exact amount of water in the jugs. For example, as we saw in class, to get 2 gallons, fill the 5 gallon jug and use those 5 gallons to fill the 3 gallon jug, which we dump out, giving 5 - 3 = 2 gallon left over in the 5 gallon jug.

Note: outside of Blackboard, the URL for this web page is http://www.math.unl.edu/~bharbourne1/M203ESpr2011/Homework1.html . Replace Homework1.html with DH3.html to see the video clip for this problem that was in the movie.