5c and 9c Stamps

The Problem

The Otehaihai Post Office only sells 5c and 9c stamps. What amounts of postage can be made up from these denominations? (The Post Office has an inexhaustible supply.)

Teaching sequence

1. Pose the problem to the class.
2. As a class list some of the values (5, 9, 5 + 9 etc).
3. Let the children work on the problem with a partner. Tell them that they should try to find a conjecture.
4. The children may become stuck. For instance they may say that “the answer is all combinations of 5 and 9, so what is there to do?” In this case, you will need to tell them that they are looking for some simple way of telling if a number can be made from 5 and 9 or not. You might suggest that they start at 1, then 2 and so on to see which values they can make.
5. If they conjecture that it looks like everything from 32 onwards can be made, ask them if they can justify this. You might need to encourage them to think about what happens when you keep adding 5 to a single number or a set of numbers.
6. Share solutions.
7. Encourage them to go on to try the Extension problem.
8. Ask them to write up what they have done. This should include a justification of what they have found.

Extension to the problem

How would things change if the stamps were 5c and 11c?
Can you guess a general result with two denominations of stamps where one denomination is 5c?

Solution

The solution method that we give here follows the same pattern as that of 3c and 5c Stamps, Number, Level 3.

A good way to start here is to experiment. For instance, make a table showing the numbers 1 to 50 and put a tick against those that can be made and a cross against those that can’t. What amounts seem to be working are 5, 9, 10, 14, 15, 18, 19, 20, 23, 24, 25, 27, 28, 29, 30, and everything from 32 onward

Now that makes a nice conjecture but how can it be justified? Can you make 32, 33, 34, 35, 36? Yes. Fine, then add 5 to each of these and you’ll get 37, 38, 39, 40, 41. But then add 5 to all of these and you’ll get 42, 43, 44, 45, 46. Can you see now that eventually you will get any number you want that is bigger than 32, simply by adding enough fours?

Other methods will work but this is probably the most efficient.

Solution to the extension

The same approach will work with 5 and 11. Here you can get 5, 10, 11, 15, 16, 20, 21, 22, 25, 26, 27, 30, 31, 32, 33, 35, and everything from 40 onwards.

Now 5 and t is a different kettle of fish. We suggest you get the students to try various values of t so that they can look for patterns and produce a conjecture. Forget about the small values you can get. They might come up with the conjecture that all numbers that are greater than or equal to 4(t – 1) can be made. That is on the track but doesn’t work if t = 55. In fact you need 5 and t to have no factors in common in order to get 4(t – 1).

Proving this last conjecture requires a bit of algebra that is past Level 4. But if you have a particularly bright group you might like to try it out. We give the full proof in What is t?, Algebra, Level 6.