# Fill and Pour

by Daniel
(Yonkers, NY, USA)

Jim has five containers:
1. A full 20 oz container
2. A full 20 oz container
3. An empty 20 oz container
4. An empty 9 oz container
5. An empty 3 oz container

He needs to fill and pour until there are:
- 10 oz in container #1
- 10 oz in container #2
- 10 oz in container #3
- 10 oz combined in containers 4 and 5

So far, I have made the following moves:
1. 2 to 4
2. 4 to 3
3. 2 to 4
4. 4 to 3
5. 1 to 4
6. 4 to 3
7. 1 to 5
8. 1 to 2
9. 4 and 5 to 1
10. 1 to 4
11. 1 to 5

This leaves me with 0 oz in container 1, 10 oz in container 2, 20 oz in container 3, 9 oz in container 4 and 1 oz in container 5. I am so stuck! I need help please!!!

### Comments for Fill and Pour

 Aug 21, 2010 Discrete Mathematics - Fill & Pour by: Staff The answer: The branch of mathematics which addresses this type of problem is called DISCRETE MATHEMATICS. The first question to ask yourself is: Is it even possible to solve this problem? The only numbers that can be used to solve this problem are specific single values – in your case, only certain integers. The volume of liquid in each container can only be a linear combination of 20, 9, and 3 oz. Every combination possible will always be a multiple of the greatest common factor. To find the answer to this question, find the greatest common factor of the three volumes stated in the problem: 20 oz, 9 oz, and 3 oz. 20 = 1 x 2 x 2 x 5 9 = 1 x 3 x 3 3 = 1 x 3 As you can see, the greatest common factor (GCF) is the number 1 (one). The target volume is 10 oz (from your statement of the problem). Can you combine the container volumes to reach the target volume? The answer is yes. You can reach the target volume of 10 oz using a linear combination of the 20, 9, and 3 oz volumes. So the problem appears to be solvable. 20 x 2 – 3 x 10 = 10, the taget amount I was not able to completely solve the problem before moving on to other things. However, it should be possible to solve it. I’m listing what I have. Take it from here: - 10 oz in container #1 - 10 oz in container #2 - 20 oz in container #3 - 0oz combined in containers 4 and 5 …………………20 oz…………..20 oz…………20 oz……………9 oz………………….3 oz …………………container………container…….container……….container……………container Fill……………..20………………20…………….0………………….0…….………………..0 Available………0………………..0……………..20………………..9………………………3 repeat 2 times: pour #1 into #4, then #4 into #3 fill………………2………………..20…………….18………………..0……………………..0 available………18………………0………………2…………………9……………………..3 pour #1 into #5 fill………………0………………..20……………..18………………..0……………………..2 available………20………………0……………….2………………....9……………………..1 pour #2 into #4 fill………………0………………11……………..18…………………9……….…………….2 available………20……………..9……………….2………………….0…………………….1 pour #2 into #5 fill………………0…………...…10……………..18…………………9…………………….3 available………20…….………10……………..2…………………..0…………………….0 pour #5 into #3 fill………………0…………..…10……………..20…………………9…………………….1 available………20……………10……………..0…………………..0…………………….2 pour #4 into #1, and #5 to #1 fill………………10……………10……………20………………….0…………………….0 available………10……………10……………0……………………9…………………….3 Thanks for writing. Staff www.solving-math-problems.com