Pour Favor

Set Up of Fill and Pour:

We have two cups. A 5 ounce cup and an 8 ounce cup. We want 4 ounces of water. How can we fill a cup with 4 ounces using only the two cups we have?

General Set Up:

We have two cups of different sizes and we want a certain amount of water/ soda. How can we fill one of the cups with that amount of liquid by only using the two cups we have?

Various Methods Used:

I only had to try this once to get it right. I started by thinking about different ways I can get the number 4 from 5 and 8. I knew if I started with a full 8 ounces and poured it into the 5 ounce cup that would leave me with 3 ounces in the 8 ounce cup, but I couldn't think of how to get only one more ounce into the cup. So I decided to start by filling the 5 ounce cup. I knew I could fill it, pour it into the 8 ounce cup, fill it again and pour it again, and I would be left with two ounces in the 5 ounce cup. So I did that. After I did it I realized that there wasn't an easy way for me to get another two ounces into the 5 ounce cup, so I just decided to empty the 8 ounce cup and pour in the two ounces I had. When I then poured another 5 ounces into the 8 ounce cup I realized that I had stumbled upon my solution. I could just fill the 5 ounce cup once more, top off the 8 ounce cup and be left with 4 ounces in my 5 ounce cup.

So I only had to go through the process first, but I did reason through my steps in my head which helped me choose a starting point and helped me follow the correct steps to the solution.

After I did the initial problem with a 5 ounce cup and an 8 ounce cup I started doing more problems that Fill and Pour offered. I noticed that, for the numbers that were being provided for me, if I followed the same procedure that I followed for the first problem, then they all worked.

General Method:

Start by filling the smaller cup and pouring it into the larger cup.
Then fill the smaller cup again and pour as much as you can into the larger cup to fill it.
Repeat until larger cup is full.
Empty the larger cup and pour the remaining amount from the smaller cup into the larger cup.
Fill the smaller cup and pour it into the larger cup.
Repeat this until you get your desired result.

The first problem is solvable!! And here's how:

Start by filling the 5 ounce cup and pour it into the 8 ounce cup.
Then fill the 5 ounce cup again and pour as much as you can into the 8 ounce cup to fill it. We end up with the 8 ounce cup full, and the 5 ounce cup with 2 ounces in it.
Now empty the 8 ounce cup.
Pour the two ounces from the 5 ounce cup into the 8 ounce cup.
Now fill the 5 ounce cup all the way and pour it into the 8 ounce cup. There are now 7 ounces in the 8 ounce cup.
Fill the 5 ounce cup again and pour as much as you can into the 8 ounce cup. You can only pour 1 ounce into the 8 ounce cup, leaving 4 ounces in the 5 ounce cup.

Here's another problem that is solvable using the general method:

Problem: We have a 7 ounce cup and an 11 ounce cup and we want 5 ounces of water

Start by filling the 7 ounce cup and pouring it into the 11 ounce cup.
Then fill the 7 ounce cup again and pour as much of it as you can into the 11 ounce cup. This will completely fill the 11 ounce cup, and leave 3 ounces in the 7 ounce cup.
Empty the 11 ounce cup and pour the remaining 3 ounces from the 7 ounce cup into the 11 ounce cup.
Fill the 7 ounce cup again and pour it into the 11 ounce cup. This will give you 10 ounces in the 11 ounce cup.
Fill the 7 ounce cup again and pour as much of it as you can into the 11 ounce cup. This will fill the 11 ounce cup, and leave 6 ounces in the 7 ounce cup.
Empty the 11 ounce cup. Pour the 6 ounces from the 7 ounce cup into the 11 ounce cup.
Fill the 7 ounce cup again and pour as much of it as you can into the 11 ounce cup. This will fill the 11 ounce cup and leave 2 ounces in the 7 ounce cup.
Empty the 11 ounce cup and pour the remaining 2 ounces from the 7 ounce cup into the 11 ounce cup.
Fill the 7 ounce cup and pour it into the 11 ounce cup.
Fill the 7 ounce cup again and pour as much of it as you can into the 11 ounce cup. This will fill the 11 ounce cup, and leave 5 ounces in the 7 ounce cup.

Solvable or Unsolvable?

For this set up, when I first started solving problems I went on faith that all the problems were solvable. And that ended up being the case for the numbers that the site provided for me. But how will I know if it is solvable or not for other numbers? When I started exploring the "Challenge" questions the solution came to me! The first challenge problem I tried started with a 6 ounce cup and a 9 ounce cup. I started with my usual procedure. I filled the 6 ounce cup, poured it into the 9 ounce cup. Filled the 6 ounce cup again, poured as much of it as I could into the 9 ounce cup, which filled the 9 ounce cup and left 3 ounces in the 6 ounce cup. I emptied the 9 ounce cup, poured the remaining 3 ounces from the 6 ounce cup into the 9 ounce cup. I filled the 6 ounce cup again and poured as much of it as I could into the 9 ounce cup which ended up filling the 9 ounce cup and emptying the 6 ounce cup. So once I emptied the 9 ounce cup I was back to the beginning with two empty cups. For this situation it is possible to fill a cup with 3, 6 or 9 ounces, but nothing else. This is the case because 6 and 9 are both multiples of 3, so adding and subtracting 6 and 9 multiple times will always give us multiples of 3, and because the largest amount we can have is 9, the only amounts we can get are 3, 6 and 9.

When the two cups share a factor, then they will only be able to reach amounts that are also a multiple of that factor. For example, one of the practice problems asks us to use an 8 ounce and 12 ounce cup to get 4 ounces. 8 and 12 are both multiples of 2 and 4, so they can get to anything that is a multiple of 2 and 4, which 4 is. So they can get 4, but they couldn't get 1,3,5,6,7,9,10 or 11 because none of those are multiples of 2 or 4.

When two numbers have a greatest common factor (GCF) of 1 then they can combine to make any number, but when two numbers have a GCF other than 1, they can only combine to get numbers that also have that factor. The Euclidean algorithm gives us the procedure for finding the GCF of two numbers.

Example:

GCF(19, 15) = ?

19= (1)x15 + 4

15= (3)x4 + 3

4= (1)x3 + 1

3= (3)x1 + 0

So the GCF of 19 and 15 is 1. When we follow the Euclidean algorithm down to the GCF, and the GCF is 1 (like in the example), then we can follow it back and see that 1= 4-3, 3=15- (3)4, and 4= 19- 15. Well, plug in the equations for 4 and 3, then 1= (19- 15) - (15- (3)4)= 31-30=1.

To find the GCF of (a, b) we follow the same procedure.

a = q₀ b + r₀

b = q₁ r₀ + r₁

r₀ = q₂ r₁ + r₂

r₁ = q₃ r₂ + r_{3 ...}

r_k₋₂ = q_k r_k₋₁ + r_k

And we could then do what we did in the Example and follow it back to find what combination (adding, subtracting and multiplying) of a and b make the GCF. GCF(a,b)= (X)a+ (Y)b. When the GCF is 1 we can then find that equation that shows us what combination of the two numbers makes 1, and then multiply it by any number to get the desired result. In relation to our fill and pour problem that means that we can continue the process of filling and pouring any number of times to get our desired result. Now, back to the Euclidean algorithm. If we have two numbers that don't have a GCF of 1, then when we find that equation GCF(a,b)= (X)a+ (Y)b, but because the GCF is not 1 we can still multiply this equation by any number, but we can only get numbers that are multiples of the GCF. In relation to our fill and pour problem that means that we can continue the process of filling and pouring and number of times, but we can only end up with amounts that are multiples of the GCF of the two cup sizes.

We can see this process from a couple of examples.

We have a 6 ounce and 10 ounce cup. Fill and pour them as needed to get 3 ounces. If we follow our general procedure from the beginning we will fill the 6 ounce cup, empty it into the 10 ounce cup. Fill the 6 ounce cup again, pour as much of it as we can into the 10 ounce cup. That will leave 2 ounces in the 6 ounce cup. Empty the 10 ounce cup and pour the remaining 2 ounces from the 6 ounce cup into the 10 ounce cup. Fill the 6 ounce cup and pour it into the 10 ounce cup. Fill the 6 ounce cup again and pour as much of it as you can into the 10 ounce cup. This will leave 4 ounces in the 6 ounce cup. Empty the 10 ounce cup and pour the remaining 4 ounces from the 6 ounce cup into the 10 ounce cup. Fill the 6 ounce cup and pour it into the 10 ounce cup. This will fill the 10 ounce cup and empty the 6 ounce cup, so when we empty the 10 ounce cup we will have two empty cups. So we were able to get the amounts 2, 4, 6, 8 and 10.

Now lets look at the Euclidean algorithm to find the GCF of 6 and 10.

GCF(6, 10)=?

10= (1)6+4

6= (1)4+ 2

4= (2)2+ 0

GCF(6, 10)= 2

So 2= 6-4, 4= 10- 6, so 2= 6- (10-6), 2= -10+(2)6. We can not multiply this equation by a whole number to get 3. So we can fill and pour the cups any amount of times to get 3 ounces.

Pedagogical Paragraph:

I completed this task by using a virtual manipulative. The task of putting 4 ounces into a cup ended up being simple, but there are a lot of mathematical concepts and lessons that can be learned from this process. For instance, to begin my investigation I didn't actually start playing with the amounts of water, I started by thinking about the relationship between the numbers 5 and 8. If students don't have a solid foundation on numbers and how they are related (subtraction and addition) this virtual manipulative is a great way to introduce that concept to them because they can see that when we add 5+3 we get 8, and if we started with two 5's, we then have a remainder of 2. Two 5's added together give us 10, so we can see that 10- 8 =2. Students can also learn deductive reasoning. For example, when I thought about starting by filling the 8 ounces cup. What could I have done from there? I could have poured it into the 5 ounce cup, leaving 3 ounces in the 8 ounce cup. Then what? poured more into the the 5 ounce cup? no, it was full. Maybe poured the 5 ounce cup into the 8 ounce cup? no, that would just be putting the water back. It is important for students to be able to think through steps and to realize what they are doing during each step.

I had to dig deeper to figure out when a problem was solvable or not. When I participated in the challenge problems, which are described above, I realized the connection the the GCF and the solvability of the problem. When the GCF is 1, then the two numbers can be added or subtracted as many time as they need to create any number. This is because they can be added or subtracted some amount of times to get one. Once you have one you just add that same process over and over until you get the desired number. When the GCF is greater than one, then the two numbers can't be added or subtracted to get one, the smallest you can get is the GCF, so they can only be combine to make numbers that are multiples of the GCF.