One of the things a math teacher needs is a supply of good problems. Occupy Math and his collaborator Andrew McEachern have coined a new term in connection with this need: Problem Factories. A problem factory has two parts. The first is a basic understanding of a mathematical fact, the second is a general type of problem or puzzle based on that fact. Ideally the understanding of the math will specify which versions of the problem can be done and give an idea of how hard they are. From this point, the way forward lies in an example.
The motivating problem for the first problem factory.
Pose the problem:
Suppose that we have two jugs. The first can hold five liters of water the second holds three. Can we measure out one liter of water? You can fill a jug, empty it out, or pour the contents of one jug into another. You cannot partially fill a jug by eye to get a measured amount of water.
The series of pictures above shows the sequence of full and partly full jugs that end with one measured liter of water in the 5-liter jug. The steps are:
Fill the five liter jug; fill the 3-liter jug from the five liter one leaving two liters behind; empty the 3-liter jug; pour the two liters into the 3-liter jug; re-fill the 5-liter jug; finish filling the 3-liter jug from the 5-liter jug, leaving 4 liters; empty the 3-liter jug; fill the 3-liter jug from the 5-liter jug, leaving one liter behind.
Done! This is a cute puzzle problem, but it is not yet a problem factory. Let’s look into the underlying math.
The greatest common divisor of two whole numbers A and B is the largest whole number that divides both A and B. We write GCD(A,B) to gives us a shorthand for “the greatest common divisor of A and B“. Here is the secret sauce: jars of size A and B can be used to measure any amount of water that is a multiple of GCD(A,B). Other amounts are not possible. This is a nifty sort of problem factory because it gives you a huge supply of problems (just choose A, B, and the target amount of water) and it lets you know ahead of time if there is a solution.
Different types of problems available from this factory.
- If the target amount of water is smaller than the larger jug, then some pouring out is needed and you get an engaging puzzle. Some pouring back-and-forth is needed.
- If you are measuring an amount of water larger than the largest bucket (into a tub or basin), then the strategy is to get close with the larger jug and finish with the small one. The problem is spotting when to shift to the smaller jug. If, for example, you wanted to measure 91 liters in a vat using the 5 and 3 liter jugs from the last example, you would empty the 5-liter jug into the vat 17 times and the 3-liter jug twice. In arithmetic: 17*5+2*3=85+6=91.
- It is possible to pose impossible problems using this problem factory. The link is to a previous Occupy Math that discusses impossible problems, both use and abuse.
- The solution to the 91 liter vat problem is not unique, e.g.: 14*5+7*3=70+21=91. Non-unique solution is an interesting topic in its own right, but it also opens another door. Make the problem a little more difficult (or make it competitive) by requesting the smallest number of moves to the solution.
- You can take an impossible problem and transform it into an approximation problem: ask for the closest possible approach to the target amount of water.
Deeper math in this factory.
A teacher will want to know where this fits into the rest of math. The problems from this problem factory are good practice in arithmetic and problem solving. Using several of these problems — especially a mix of possible and impossible problems with the same jugs — can be structured as a discovery exercise that will permit students to evolve the concept of a greatest common divisor. Why do we want to go there?
Greatest common divisors are what you remove from the numerator and denominator of a fraction to put it in simplest form. Occupy Math has noted that, in some ways, education about fraction arithmetic is on life support. The problem factory outlined in this blog gives you a way to approach simplification of fractions from left field and help students develop their critical number sense.
Which problems are hard?
The problems where you are filling a large external container are easier than the ones with a lot of filling and pouring out to get a small amount of water in one of the jugs. There is a wonderful technique for computing the GCD of two numbers; it is called the Euclidean algorithm. Solutions to jug problems often parallel the Euclidean algorithm, but the math on which the Euclidean algorithm is based gives us an interesting fact. For numbers of roughly the same size, pairs of consecutive Fibonacci numbers take the most steps using the Euclidian algorithm — and can let you pour any whole number quantity of water.
Remember that the Fibonacci numbers are obtained by adding the last two to get the next. Starting with a couple of ones we get: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,… This means that using an 8-ounce and 13-ounce cup would give you longer (and probably harder) problems than an 8-ounce and a 12-ounce cup.
While it was not written up as a problem factory, the “can you draw this” type of problem is easy to see as a problem factory. The instructor can rapidly create problems and know — via a simple test — if they can be solved or not.
Occupy Math hopes you’ve enjoyed this problem factory — and would love it if you get ideas for your own problem factories. Occupy Math and his team are writing an article on problem factories and intend to bring out a book of problems. So far we’ve found 12 problem factories (including the two in this post). If we adopt one of your ideas we will send you a copy of the book — unless you want to do your own book, of course. Cool ideas? Please comment or tweet.
I hope to see you here again,
University of Guelph,
Department of Mathematics and Statistics