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.)

What is this problem about?
This problem involves the use of sums and multiples of 3 and 5.  It also involves understanding that if you have 3 consecutive numbers then you can produce all subsequent numbers by adding multiples of 3 to those numbers. For example: 13, 14, 15 adding three to each gives 16, 17, 18 and then adding 3 to these gives 19, 20, 21 and so on.

Achievement Objectives
Number (Level 4)
- make sensible estimates and check the reasonableness of answers

Mathematical Processes
- devise and use problem solving strategies (guess and check, make a table, look for patterns)

Resources
stamps/ envelopes to pose the problem
Blackline master of the problem (English)
Blackline master of the problem (Maaori)

Specific learning outcomes
The children will be able to:
- describe what it means for a sequence to carry on infinitely
- look for patterns using multiples

1. Pose the problem to the class.
2. As a class list some of the values (3, 5, 3 + 5 etc)
3. Let the children work on the problem with a partner.
4. Then if the children are stuck you may suggest starting at 1, then 2 and so on to see which values they can make.
5. As the children work encourage them to think about what happens when you keep adding 3 to a single number or a set of numbers.
6. Share solutions.

Extension to the problem
How would things change if the stamps were 3c and 7c?
Can you guess the result for 3-cent stamps and s-cent stamps, where s is some whole number?

Solution to the problem
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. 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, 10? Yes. Fine, then add 3 to each of these and you’ll get 11, 12, 13. But then add 3 to all of these and you’ll get 14, 15, 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.)

Extension: The same approach will work with 3 and 7. Here you can get 3, 6, 7, 9, 10, and everything from12 onwards.

Now 3 and s is a different kettle of fish. We suggest you get the children to try various values of s 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 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 2(s–1).