3c and 5c Stamps

The Problem

The Otehaihai Post Office only sells 3c and 5c 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 (3, 5, 3 + 5 etc). You might like to use actual stamps (or stamps that you have made) to get them started.
3. Let the students work on the problem with a partner.
4. The students may become stuck. For instance they may say that “the answer is all combinations of 3 and 5, 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 a combination of 3 and 5 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 can see that it looks like everything from 8 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 3 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 3c and 7c?

Can you guess a general result with two denominations of stamps where one denomination is 3c?

Solution

A good way to start here is to experiment. For instance, make a table showing the numbers 1 to 20 and put a tick against those that can be made and a cross against those that can’t. (On the other hand, you might like to make a table like Sara’s Table, Algebra, Level 3.) What amounts seem to be working are 3, 5, 6, and everything from 8 onwards.

Now that is a nice conjecture but how can it be justified? Can you make 8, 9, and 10? Yes. Fine, then add 3 to each of these and you’ll get 11, 12, and 13. But then add 3 to all of these and you’ll get 14, 15, and 16. Can you see now that eventually you will get any number you want that is bigger than 8, simply by adding enough threes?

(Alternately, though a little longer, show that 8, 9, 10, 11, 12 can be done and add fives to get to any number above 8 that you want.)

Solution to the Extension:

The same approach will work with 3 and 7. Here you can get 3, 6, 7, 9, 10, and everything from 12 onwards.

Now 3 and s is a different kettle of fish. We suggest that you get the students to try various values of s so that they can look for patterns and produce a conjecture. (Forget about the small values that you can get.) They might come up with the conjecture that you can get all numbers from 2(s – 1) onwards. That is on the track but doesn’t work if s = 9. In fact you need 3 and s to have no factors in common in order to get everything from 2(s – 1) onwards.

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 s?, Level 6.